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;
using ll = long long;

bool solve() {
    ll N, sa = 0;
    cin >> N;
    vector<ll> A(N);
    for ( int i = 0; i < N; i++ ) {
        cin >> A[i];
        sa += A[i];
    }
    ll nn = N*(N+1)/2;
    if ( sa % nn != 0 ) return false;
    ll k = sa / nn;
    for ( int i = 0; i < N; i++ ) {
        ll d = - ( A[(i+1)%N] - A[i] - k );
        if ( d < 0 || d % N != 0 ) return false;
    }
    return true;
}

int main() {
    string ans = solve() ? "YES" : "NO";
    cout << ans << "\n";
    return 0;
}