A Communist regime is trying to redistribute wealth in a village. They have have decided to sit everyone around a circular table. First, everyone has converted all of their properties to coins of equal value, such that the total number of coins is divisible by the number of people in the village. Finally, each person gives a number of coins to the person on his right and a number coins to the person on his left, such that in the end, everyone has the same number of coins. Given the number of coins of each person, compute the minimum number of coins that must be transferred using this method so that everyone has the same number of coins.

Input

There is a number of inputs. Each input begins with n (n < 1000001), the number of people in the village. n lines follow, giving the number of coins of each person in the village, in counterclockwise order around the table. The total number of coins will fit inside an unsigned 64 bit integer.

Output

For each input, output the minimum number of coins that must be transferred on a single line.

Sample Input

3

100

100

100

4

1

2

5

4

Sample Output

0

4

算法竞赛入门经典训练指南上有题意与分析，不懂的可以去翻阅。

 //Asimple
#include <bits/stdc++.h>
#define INF (1<<20)
#define mod 10007
#define swap(a,b,t) t = a, a = b, b = t
#define CLS(a, v) memset(a, v, sizeof(a))
#define debug(a)  cout << #a << " = "  << a <<endl
using namespace std;
typedef long long ll;
const int maxn = 1000001;
ll n, m, res, ans, len, T, k, num, sum;
ll x, y;
ll a[maxn], c[maxn];

/*
假设 An表示n个人原有的金币数
xn表示第n个人给n-1个人xn枚金币（x1表示1给n）
M表示最终每个人拥有的金币数 
则可推出通式： xn = x1-Cn 其中 Cn = A1+A2+...+An-n*M
由于要求是最小的金币数，所以x1应当取C数组的中位数 
*/

void input() {
    ios_base::sync_with_stdio(false);
    while( cin >> n ) {
        sum = 0;
        for(int i=1; i<=n; i++) {
            cin >> a[i];
            sum += a[i];
        }
        ll M = sum / n;
        c[0] = 0;
        for(int i=1; i<n; i++)
            c[i] = c[i-1]+a[i]-M;
        sort(c, c+n);
        ll x1 = c[n/2];
        ans = 0;
        for(int i=0; i<n; i++)
            ans += abs(x1-c[i]);
        cout << ans << endl;
    }
}

int main(){
    input();
    return 0;
}