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 rep(i,n) for (int (i)=(0);(i)<(int)(n);++(i))

int main() {
    long long N, sum = 0; cin >> N;
    vector<long long> A(N), B(N);
    rep(i,N) {
        cin >> A[i];
        sum += A[i];
    }

    long long t = 1LL * N * (N + 1) / 2;
    if (sum % t) {
        puts("NO");
        return 0;
    }

    long long cnt = sum / t;
    rep(i,N) {
        B[i] = A[(i+1)%N] - A[i] - cnt;
    }
    bool ok = true;
    rep(i,N) {
        if (B[i] > 0 || (-B[i]) % N) ok = false;
    }

    cout << (ok ? "YES" : "NO") << endl;

    return 0;
}
