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 <cstdio>
using namespace std;

long long int N, A[100010];
int main() {
    scanf("%lld", &N);
    long long int sum = 0, div = N*(N+1)/2;
    for(int i=0; i<N; i++) {
        scanf("%lld", &A[i]);
        sum += A[i];
    }

    bool ok = (sum % div == 0);
    long long int sum_op = sum / div;
    for(int i=0; i<N; i++) {
        long long int val = A[i] - A[(i+1)%N] + sum_op;
        ok &= (val >= 0 && val % N == 0);
    }
    printf("%s\n", ok ? "YES" : "NO");
    return 0;
}