Minimize the error CodeForces

You are given two arrays A and B, each of size n. The error, E, between these two arrays is defined . You have to perform exactly k₁ operations on array A and exactly k₂ operations on array B. In one operation, you have to choose one element of the array and increase or decrease it by 1.

Output the minimum possible value of error after k₁ operations on array A and k₂operations on array B have been performed.

Input

The first line contains three space-separated integers n (1 ≤ n ≤ 10³), k₁ and k₂ (0 ≤ k₁ + k₂ ≤ 10³, k₁ and k₂ are non-negative) — size of arrays and number of operations to perform on A and B respectively.

Second line contains n space separated integers a₁, a₂, ..., a_n ( - 10⁶ ≤ a_i ≤ 10⁶) — array A.

Third line contains n space separated integers b₁, b₂, ..., b_n ( - 10⁶ ≤ b_i ≤ 10⁶)— array B.

Output

Output a single integer — the minimum possible value of  after doing exactly k₁ operations on array A and exactly k₂ operations on array B.

Examples

Input

2 0 0
1 2
2 3

Output

2

Input

2 1 0
1 2
2 2

Output

0

Input

2 5 7
3 4
14 4

Output

1

Note

In the first sample case, we cannot perform any operations on A or B. Therefore the minimum possible error E = (1 - 2)² + (2 - 3)² = 2.

In the second sample case, we are required to perform exactly one operation on A. In order to minimize error, we increment the first element of A by 1. Now, A = [2, 2]. The error is now E = (2 - 2)² + (2 - 2)² = 0. This is the minimum possible error obtainable.

In the third sample case, we can increase the first element of A to 8, using the all of the 5 moves available to us. Also, the first element of B can be reduced to 8using the 6 of the 7 available moves. Now A = [8, 4] and B = [8, 4]. The error is now E = (8 - 8)² + (4 - 4)² = 0, but we are still left with 1 move for array B. Increasing the second element of B to 5 using the left move, we get B = [8, 5] and E = (8 - 8)² + (4 - 5)² = 1.

【题目概述】

给你两个数字序列a，b，每个序列长度都为n，然后E=∑（a[i]-b[i]）^2。现在你可以改变a，b序列中的元素k1次和k2次，每次可以使一个元素加一或者减一。使得改变结束之后E的值最小

【思路阐述】

每一次的操作都会使差值变化，+1或-1，目标是使差值离0越近越好，那么当存在有大于0的差值时，就让改差值减一，如果当所有差值为0的时，就让其中一个（就第一个）差值加一。

#include<bits/stdc++.h>
using namespace std;
struct node{
    int a;
    int b;
    int dif;
}num[1009];

bool cmp(node a,node b) {
    return a.dif > b.dif;
}

int main() {
    int n,k1,k2;
    while(~scanf("%d %d %d",&n,&k1,&k2)) {
        for(int i = 0; i < n; i++) scanf("%d",&num[i].a);
        for(int i = 0; i < n; i++) {
            scanf("%d",&num[i].b);
            num[i].dif = abs(num[i].a - num[i].b);
        }
        sort(num,num+n,cmp);
        int op = k1 + k2;
        int count = 0;
        for(int i = 0; i < op; i++) {
            if(num[0].dif == 0) num[0].dif++;
            else num[0].dif--;
            count++;
            sort(num,num+n,cmp);
        }
        long long int ans = 0;
        for(int i = 0; i < n; i++) {
            ans += pow(num[i].dif,2);
        }
        
        cout<<ans<<endl;
    }
    return 0;
}