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>
using namespace std;
typedef long long ll;
ll n, a[100100], sum, res;
bool ans = true;
int main(void){
    // Your code here!
    cin >> n;
    for (ll i = 1; i <= n; i++) {
        cin >> a[i];
        sum += a[i];
    }
    a[0] = a[n];
    if (sum % ((1 + n) * n / 2) != 0) ans = false;
    sum /= ((1 + n) * n / 2);
    for (ll i = 0; i < n; i++){
        if ((sum - (a[i + 1] - a[i])) % n != 0 || sum < a[i + 1] - a[i]) ans = false;
        res += (sum - (a[i + 1] - a[i])) / n;
    }
    if (res != sum) ans = false;
    cout << (ans ? "YES" : "NO") << endl;
}
