Country of Metropolia is holding Olympiad of Metrpolises soon. It mean that all jury members of the olympiad should meet together in Metropolis (the capital of the country) for the problem preparation process.
There are *n* + 1 cities consecutively numbered from 0 to *n*. City 0 is Metropolis that is the meeting point for all jury members. For each city from 1 to *n* there is exactly one jury member living there. Olympiad preparation is a long and demanding process that requires *k* days of work. For all of these *k* days each of the *n* jury members should be present in Metropolis to be able to work on problems.
You know the flight schedule in the country (jury members consider themselves important enough to only use flights for transportation). All flights in Metropolia are either going to Metropolis or out of Metropolis. There are no night flights in Metropolia, or in the other words, plane always takes off at the same day it arrives. On his arrival day and departure day jury member is not able to discuss the olympiad. All flights in Megapolia depart and arrive at the same day.
Gather everybody for *k* days in the capital is a hard objective, doing that while spending the minimum possible money is even harder. Nevertheless, your task is to arrange the cheapest way to bring all of the jury members to Metrpolis, so that they can work together for *k* days and then send them back to their home cities. Cost of the arrangement is defined as a total cost of tickets for all used flights. It is allowed for jury member to stay in Metropolis for more than *k* days.
Input
The first line of input contains three integers *n*, *m* and *k* (1 ≤ *n* ≤ 105, 0 ≤ *m* ≤ 105, 1 ≤ *k* ≤ 106).
The *i*-th of the following *m* lines contains the description of the *i*-th flight defined by four integers *d**i*, *f**i*, *t**i* and *c**i* (1 ≤ *d**i* ≤ 106, 0 ≤ *f**i* ≤ *n*, 0 ≤ *t**i* ≤ *n*, 1 ≤ *c**i* ≤ 106, exactly one of *f**i* and *t**i* equals zero), the day of departure (and arrival), the departure city, the arrival city and the ticket cost.
Output
Output the only integer that is the minimum cost of gathering all jury members in city 0 for *k* days and then sending them back to their home cities.
If it is impossible to gather everybody in Metropolis for *k* days and then send them back to their home cities, output "-1" (without the quotes).
Examples
Input
```
2 6 5
1 1 0 5000
3 2 0 5500
2 2 0 6000
15 0 2 9000
9 0 1 7000
8 0 2 6500
```
Output
```
24500
```
Input
```
2 4 5
1 2 0 5000
2 1 0 4500
2 1 0 3000
8 0 1 6000
```
Output
```
-1
```
Note
The optimal way to gather everybody in Metropolis in the first sample test is to use flights that take place on days 1, 2, 8 and 9. The only alternative option is to send jury member from second city back home on day 15, that would cost 2500 more.
In the second sample it is impossible to send jury member from city 2 back home from Metropolis.
## 题意：
给你N个大使，M个航班，和一个天数K，
对于每一个航班i，有四个信息，分别是日期，起始站，目标站，价格。
其中起始站和目标站一定有一个是0节点。
让你把1~n个大使都从第i个城市接到0节点，开K天及以上的的会议，然后再全部送回他们的城市。
使之花费的成本最小。
## 思路：
读入的时候找出最大的天数maxtime。
从1到maxtime维护一个全部人可以送到0城市的最小消费
然后从maxtime到1反向维护一个把全部人送回去的最小消费。
然后1到maxtime 取一个min既是ans。
``` cpp
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d
",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 2000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
struct node
{
int time;
int f;
int t;
ll cost;
} a[maxn];
int n;
int m;
int k;
bool cmp(node aa, node bb)
{
return aa.time < bb.time;
}
int incnt = 0;
int outcnt = 0;
ll in[maxn];
ll out[maxn];
ll sumin[maxn];
ll sumout[maxn];
ll sum1 = 0ll;
ll sum2 = 0ll;
int main()
{
// freopen("D:\common_text\code_stream\in.txt","r",stdin);
//freopen("D:common_textcode_streamout.txt","w",stdout);
gbtb;
cin >> n >> m >> k;
ll mtime = 0;
repd(i, 1, m)
{
cin >> a[i].time >> a[i].f >> a[i].t >> a[i].cost;
mtime = max(mtime, 1ll * a[i].time);
}
sort(a + 1, a + 1 + m, cmp);
repd(i, 1, m)
{
if (a[i].t == 0)
{
if (in[a[i].f] == 0)
{
incnt++;
in[a[i].f] = a[i].cost;
sum1 += a[i].cost;
} else
{
if (in[a[i].f] > a[i].cost)
{
sum1 -= in[a[i].f];
sum1 += a[i].cost;
in[a[i].f] = a[i].cost;
}
}
if (incnt == n)
{
sumin[a[i].time] = sum1;
}
}
}
for (int i = m; i >= 1; i--)
{
if (a[i].f == 0)
{
if (out[a[i].t] == 0)
{
outcnt++;
out[a[i].t] = a[i].cost;
sum2 += a[i].cost;
} else
{
if (out[a[i].t] > a[i].cost)
{
sum2 -= out[a[i].t];
sum2 += a[i].cost;
out[a[i].t] = a[i].cost;
}
}
if (outcnt == n)
{
sumout[a[i].time] = sum2;
}
}
}
ll ans = 11731173111173111;
repd(i, 1, mtime)
{
if (sumin[i] == 0)
sumin[i] = sumin[i - 1];
else if (sumin[i - 1])
sumin[i] = min(sumin[i], sumin[i - 1]);
}
for (int i = mtime; i >= 1; i--)
{
if (sumout[i] == 0)
{
sumout[i] = sumout[i + 1];
}
else if (sumout[i + 1])
sumout[i] = min(sumout[i], sumout[i + 1]);
}
repd(i, 1, mtime - k - 1)
{
if (sumout[i + k + 1] && sumin[i])
ans = min(ans, sumout[i + k + 1] + sumin[i]);
}
if (ans != 11731173111173111)
cout << ans << endl;
else
cout << -1 << endl;
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '
');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}
```