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 <iostream>
#define REP(i, a, n) for(ll i = ((ll) a); i < ((ll) n); i++)
using namespace std;
typedef long long ll;

ll N, A[100000];

int main(void) {
  cin >> N;
  REP(i, 0, N) cin >> A[i];

  ll sa = 0;
  REP(i, 0, N) sa += A[i];

  if(sa % (N * (N + 1) / 2) != 0) {
    cout << "NO" << endl;
  } else {
    ll k = sa / (N * (N + 1) / 2);
    ll cnt = 0;
    bool f = 1;
    REP(i, 0, N) {
      ll x = (-A[(i + 1) % N] + A[i] + k);
      f = f && x >= 0 && x % N == 0;
      cnt += x / N;
    }
    cout << (f && cnt == k ? "YES" : "NO") << endl;
  }
}
