Codeforces Round #402 (Div. 2) --- C. Dishonest Sellers

Igor found out discounts in a shop and decided to buy n items. Discounts at the store will last for a week and Igor knows about each item that its price now is a_i, and after a week of discounts its price will be b_i.

Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.

Igor decided that buy at least k of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all n items.

Input

In the first line there are two positive integer numbers n and k (1 ≤ n ≤ 2·10⁵, 0 ≤ k ≤ n) — total number of items to buy and minimal number of items Igor wants to by right now.

The second line contains sequence of integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁴) — prices of items during discounts (i.e. right now).

The third line contains sequence of integers b₁, b₂, ..., b_n (1 ≤ b_i ≤ 10⁴) — prices of items after discounts (i.e. after a week).

Output

Print the minimal amount of money Igor will spend to buy all n items. Remember, he should buy at least k items right now.

Examples

input

3 1
 5 4 6
 3 1 5

output

input

5 3
 3 4 7 10 3
 4 5 5 12 5

output

Note

In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.

In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25

题意：你需要买n件商品，给出降价前和降价后（降价后的价格可能比降价前还要高...）的商品价格，给你一个数k，你在降价前至少买k件，问怎么买花费最少

感觉就是个贪心的思想，分几种情况，降价后价格比降价前还要高的话，那肯定买降价前的，买完这种之后判断你当前买的数量是否超过k，超过了就直接买降价后的

，没超过的话在降价前的一批商品中挑便宜的优先；如果不存在降价后价格比降价前还要高的情况，就排序降价前和降价后的差值，按差值从小到大买降价前，买满k个

之后加上降价后的...

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ss;
const int maxn = 2e5 + 5;
int bef[maxn], af[maxn];
struct node{
    int bef, af, chazhi;
};
bool cmp(node a, node b){
    if(a.chazhi == b.chazhi)    return a.bef < b.bef;    
    else return a.chazhi < b.chazhi;
}
node chazhi[maxn];
int n, k;

int main(){
    ss sum_pri;
    int cnt, l, cz;
    while(scanf("%d%d", &n, &k)!=EOF && n){
        sum_pri = cnt = 0;
        for(int i=0; i<n; i++){
            scanf("%d", &bef[i]);
        }
        for(int i=0; i<n; i++){
            scanf("%d", &af[i]);
        }
        l=0, cz;
        for(int i=0; i<n; i++){
            cz = bef[i] - af[i];
            if(cz < 0){
                sum_pri += bef[i];
                cnt++;
            }else{
                chazhi[l].bef = bef[i];
                chazhi[l].af = af[i];
                chazhi[l].chazhi = cz;
                l++;
            }
        }
        if(cnt>=k){
            for(int i=0; i<l; i++){
                sum_pri += chazhi[i].af;
            }
            cout << sum_pri << endl;
        }
        else{
            int i;
            sort(chazhi, chazhi+l, cmp);
            for(i=0; i<l; i++){
                sum_pri += chazhi[i].bef;
//                cout << chazhi[i].bef << " ";
                cnt++;
                if(cnt>=k) break;
            }
            i++;
            for(i; i<l; i++){
                sum_pri += chazhi[i].af;
    //            cout << chazhi[i].af << " ";
            }
            cout << sum_pri << endl;
        }
    }
    
    
    return 0;
}