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 debug(x) cout << #x" = " << x;

#define st first
#define nd second

using namespace std;
using namespace placeholders;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;

const int MAXN = 1E5 + 10;

ll a[MAXN], b[MAXN];

int main(){
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; ++i)
		scanf("%lld", a + i);
	ll s = accumulate(a, a + n, 0ll);
	ll n0 = n * (n + 1ll) / 2;
	if (n == 1)
		puts("YES");
	else if (s % n0)
		puts("NO");
	else{
		ll m = s / n0;
		a[n] = a[0];
		bool flag = true;
		for (int i = 0; i < n; ++i){
			ll t = m - (a[i + 1] - a[i]);
			b[i] = t / n;
			flag &= 0 <= b[i] && b[i] <= m && t % n == 0;
		}
		puts(flag && accumulate(b, b + n, 0ll) == m ? "YES" : "NO");
	}
	
	return 0;
}