B. Increase and Decrease

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarpus has an array, consisting of n integers a₁, a₂, ..., a_n. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:

• he chooses two elements of the array a_i, a_j (i ≠ j);
• he simultaneously increases number a_i by 1 and decreases number a_j by 1, that is, executes a_i = a_i + 1 and a_j = a_j - 1.

The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.

Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the array size. The second line contains space-separated integersa₁, a₂, ..., a_n (|a_i| ≤ 10⁴) — the original array.

Output

Print a single integer — the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.

Sample test(s)

input

22 1

output

1

input

31 4 1

output

3

解题说明：考虑到一对加减运算其实是数字之间大小的转换，此题其实只需要判断所有数字之和能否被n整除即可，如果能被整除就输出n，当不能被整除时我们可以让余数作为一个单独的数字，此时相同的数共有n-1个，输出n-1

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <algorithm>using namespace std;int main() {int n,i,a;int sum;sum=0;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a);sum+=a;}if(sum%n==0){printf("%d\n",n);}else{printf("%d\n",n-1);}return 0;}