Problem:
Score : 500 points

Problem Statement

There are N boxes arranged in a circle. The i-th box contains A_i stones.

Determine whether it is possible to remove all the stones from the boxes by repeatedly performing the following operation:

Select one box. Let the box be the i-th box. Then, for each j from 1 through N, remove exactly j stones from the (i+j)-th box. Here, the (N+k)-th box is identified with the k-th box.

Note that the operation cannot be performed if there is a box that does not contain enough number of stones to be removed.

Constraints

1 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

If it is possible to remove all the stones from the boxes, print YES. Otherwise, print NO.

Sample Input 1

5
4 5 1 2 3

Sample Output 1

YES

All the stones can be removed in one operation by selecting the second box.

Sample Input 2

5
6 9 12 10 8

Sample Output 2

YES

Sample Input 3

4
1 2 3 1

Sample Output 3

NO

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int compare(const void *a, const void *b) {
    return *(int*)a - *(int*)b;
}

int main(void) {
	int i,n,isSame;
	long long int cnt;
	int a[100000] = {0};
	
	scanf("%d",&n);
	
	cnt = 0LL;
	for(i = 0;i < n;i++) {
		scanf("%d",&a[i]);
		cnt += a[i];
	}
	qsort(a,n,sizeof(int),compare);
	
	isSame = 0;
	for (i = 0;i < n-1;i++) {
		if (a[i] == a[i+1]) {
			isSame = 1;
			break;
		}
	}
	
	long long int tmp = n;
	tmp *= n+1LL;
	tmp /= 2LL;
	if (cnt % tmp || a[n-1] < n || isSame) {
		printf("NO\n");
	} else {
		printf("YES\n");
	}
	
	return 0;
}