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;
typedef long long ll;

int main(void) {
  ll n;
  cin >> n;

  vector<ll> a(n);
  ll sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }

  if (sum % (n * (n + 1) / 2)) {
    cout << "NO\n";
  } else {
    ll cnt = sum / (n * (n + 1) / 2), cnt2 = 0;
    vector<ll> b = a;
    for (int i = 0; i < n; i++) {
      b[i] = a[i] - a[(i - 1 + n) % n];
      cnt2 += abs(b[i] - cnt) / n;
    }
    cout << (cnt == cnt2 ? "YES" : "NO") << endl;
  }

  return 0;
}