Educational Codeforces Round 3 C. Load Balancing

C. Load Balancing

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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：

不描述了，好难描述啊。。。

思路：

直接求平均数，大于平均数的减，小于的加，最后得到的最大最小数只差必为0或1。而sum%n代表的就是差为1的部分；然后加一下所耗的步数就可以了。

实现代码：

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;

int main()
{
    int n,i,a[],ans=,sum=;
    cin>>n;
    for(i=;i<=n;i++){
        cin>>a[i];
        sum+=a[i];
    }
    sort(a+,a+n+);
    int p = sum/n;
    int n1 = n-sum%n;
    for(i=;i<=n1;i++){
        if(a[i]>=p)
        break;
        ans+=p-a[i];
    }
    for(i=n1+;i<=n;i++){
        if(a[i]>p)
        break;
        ans+=p+-a[i];}
    cout<<ans<<endl;
}