A - Load Balancing CodeForces - 609C

A - Load Balancing

CodeForces - 609C

In the school computer room there are n servers which are responsible for processing several computing tasks. You know the number of scheduled tasks for each server: there are m_i tasks assigned to the i-th server.

In order to balance the load for each server, you want to reassign some tasks to make the difference between the most loaded server and the least loaded server as small as possible. In other words you want to minimize expression m_a - m_b, where a is the most loaded server and b is the least loaded one.

In one second you can reassign a single task. Thus in one second you can choose any pair of servers and move a single task from one server to another.

Write a program to find the minimum number of seconds needed to balance the load of servers.

Input

The first line contains positive number n (1 ≤ n ≤ 10⁵) — the number of the servers.

The second line contains the sequence of non-negative integers m₁, m₂, ..., m_n (0 ≤ m_i ≤ 2·10⁴), where m_i is the number of tasks assigned to the i-th server.

Output

Print the minimum number of seconds required to balance the load.

Example

Input

2
1 6

Output

2

Input

7
10 11 10 11 10 11 11

Output

0

Input

5
1 2 3 4 5

Output

3

Note

In the first example two seconds are needed. In each second, a single task from server #2 should be moved to server #1. After two seconds there should be 3 tasks on server #1 and 4 tasks on server #2.

In the second example the load is already balanced.

A possible sequence of task movements for the third example is:

1. move a task from server #4 to server #1 (the sequence m becomes: 2 2 3 3 5);
2. then move task from server #5 to server #1 (the sequence m becomes: 3 2 3 3 4);
3. then move task from server #5 to server #2 (the sequence m becomes: 3 3 3 3 3).

The above sequence is one of several possible ways to balance the load of servers in three seconds.

题意：有n台电脑，他们的性能分别是x1,x2...xn，高性能电脑可以每秒将自己的性能传递给低性能电脑，
即自己-1，低性能+1.问最少需要多少秒才能使它们达到平衡。（也就是它们之间的差距最大为1）

思路：先排序，求和为sum，sum/n即为最后到达平衡的最小的值，sum%n为达到平衡后为sum/n+1的数量。
将初始值中最后sum%n个数自减，这样就可以使每个数最后都变为sum/n。求和计算即可，最后记得/2.（因为
1s内，高性能-1，底性能+1）

code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
int main(){
    int n;
    vector<int> a;
    int sum = 0;
    scanf("%d",&n);
    for(int i = 0; i < n; i++){
        int x;
        scanf("%d",&x);
        a.push_back(x);
        sum += x;
    }
    int aver = sum / n;
    int mod = sum % n;
    sort(a.begin(),a.end());
    for(int i = n-1; i >= n - mod; i--){
        a[i]--;
    }
    int cnt = 0;
    for(int i = 0; i < n; i++){
        cnt += abs(a[i]-aver);
    }
    cnt /= 2;
    printf("%d",cnt);
    return 0;
}