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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define MAXN (100010)
#define ll long long

ll n;
ll a[MAXN], sum;

int main(){
	scanf("%lld", &n);
	for(int i = 1; i <= n; ++ i){
		scanf("%lld", a + i);
		sum += a[i];
	}
	ll tmp = n * (n + 1) / 2;
	if(sum % tmp) return printf("NO\n"), 0;
	ll k = sum / tmp;
	for(int i = 1; i <= n; ++ i){
		ll di = a[i % n + 1] - a[i];
		if(k < di || (k - di) % n) return printf("NO\n"), 0;
	}
	printf("YES\n");
	return 0;
}