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`.

Examples

Input

5
4 5 1 2 3


Output

YES


Input

5
6 9 12 10 8


Output

YES


Input

4
1 2 3 1


Output

NO
// わーい考察！あかり考察だーいすき！
// 嘘、嫌い
#include <iostream>
#define REP(i,n) for(int i=0;i<(n);++i)

using namespace std;

typedef long long ll;
int n, a[100000];

bool solve(){
	cin >> n;
	ll sum=0;
	REP(i, n){
		cin >> a[i];
		sum += a[i];
	}
	if(sum % ((ll)n * (n+1) / 2) != 0ll) return false;
	int k = sum / ((ll)n * (n+1) / 2); // 操作回数
	REP(i, n){
		int now = a[i], next = a[(i+1)%n];
		if(now - next < -k) return false;
		if((k + now - next) % n != 0) return false;
	}
	return true;
}

int main() {
	ios::sync_with_stdio(false);cin.tie(0);
	cout << (solve() ? "YES" : "NO") << endl;
}
