Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Office Keys（思维）

There are n people and k keys on a straight line. Every person wants to get to the office which is located on the line as well. To do that, he needs to reach some point with a key, take the key and then go to the office. Once a key is taken by somebody, it couldn't be taken by anybody else.

You are to determine the minimum time needed for all n people to get to the office with keys. Assume that people move a unit distance per 1 second. If two people reach a key at the same time, only one of them can take the key. A person can pass through a point with a key without taking it.

Input

The first line contains three integers n, k and p (1 ≤ n ≤ 1 000, n ≤ k ≤ 2 000, 1 ≤ p ≤ 109) — the number of people, the number of keys and the office location.

The second line contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ 109) — positions in which people are located initially. The positions are given in arbitrary order.

The third line contains k distinct integers b1, b2, ..., bk (1 ≤ bj ≤ 109) — positions of the keys. The positions are given in arbitrary order.

Note that there can't be more than one person or more than one key in the same point. A person and a key can be located in the same point.

Output

Print the minimum time (in seconds) needed for all n to reach the office with keys.

Examples

input

2 4 50
20 100
60 10 40 80

output

input

1 2 10
11
15 7

output

Note

In the first example the person located at point 20 should take the key located at point 40 and go with it to the office located at point 50. He spends 30 seconds. The person located at point 100 can take the key located at point 80 and go to the office with it. He spends 50seconds. Thus, after 50 seconds everybody is in office with keys.

【题意】一条坐标轴上有n个人，k把钥匙，办公室在p点。要求每个人拿一把钥匙走到办公室，问最长的距离最小是多少？

【分析】有一点得想到，将钥匙坐标从小到大排个序后，取连续的n把钥匙，肯定是最优的，现在问题是从哪个先取，枚举咯...

#include <bits/stdc++.h>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int N=1e3+50;
using namespace std;
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,p;
int a[N],b[2*N];

int main(){
    scanf("%d%d%d",&n,&m,&p);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    for(int i=1;i<=m;i++){
        scanf("%d",&b[i]);
    }
    int ans=2e9;
    sort(a+1,a+1+n);
    sort(b+1,b+1+m);
    for(int j=1;j<=m-n+1;j++){
        int s=0;
        for(int i=1;i<=n;i++){
            s=max(s,abs(a[i]-b[i+j-1])+abs(b[i+j-1]-p));
        }
        ans=min(ans,s);
    }
    printf("%d
",ans);
    return 0;
}