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<set>
#include<cstdio>
#include<algorithm>
using namespace std;

#define MAXN 100000
int N;
int A[MAXN+5];

int main(){
    scanf("%d",&N);
    long long Sum=0;
    for(int i=1;i<=N;i++){
        scanf("%d",&A[i]);
        Sum+=A[i];
    }
    long long All=1ll*N*(N+1)/2;
    if(Sum%All)
        return !puts("NO");
    int k=Sum/All;
    for(int i=1;i<=N;i++){
        int d=A[i]-A[i-1==0?N:i-1];
        if(d-k>0||(k-d)%N!=0)
            return !puts("NO");
    }
    puts("YES");
}
