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
#include <bits/stdc++.h>
using namespace std;
#define LL long long
int n;
#define N 200000
LL ai[N];
LL sum;
int main()
{
    cin >> n;
    for (int i = 1; i <= n; ++ i)
        cin >> ai[i], sum += ai[i];
    if (sum % (1ll * n * (n + 1) / 2)) {puts("NO"); return 0;}
    sum /= 1ll * n * (n + 1) / 2;
    ai[0] = ai[n];
    for (int i = 1; i <= n; ++ i)
    {
        if ((sum - (ai[i] - ai[i - 1])) % n) {puts("NO"); return 0;}
        LL d = (sum - (ai[i] - ai[i - 1])) / n;
        if (0 <= d && d <= sum); else {puts("NO"); return 0;}
    }
    puts("YES");
}
