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 = int64_t;
int main(){
    ll N;
    cin >> N;
    ll a[N];
    ll po = N * (N + 1) / 2;
    ll sum = 0;
    bool ok = 1;
    for(ll i = 0; i < N; i++){
        cin >> a[i];
        sum += a[i];
    }
    ll sa;
    if(sum % po) ok = 0;
    ll kaisu = sum / po;
    ll yo = kaisu;
    for(ll i = 0; i < N; i++){
        sa = a[(i + 1) % N] - a[i] - kaisu;
        if(sa < 0){
            if(-sa % N) ok = 0;
            yo -= -sa / N;
        }
    }
    if(yo) ok = 0;
    cout << (ok ? "YES\n" : "NO\n");
    return 0;
}