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>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=200010;
long long A[N],B[N],s; 
int n;
void solve(){
    s=0;
    for(int i=0;i<n;i++)scanf("%lld",&A[i]),s+=A[i];
    for(int i=0;i<n-1;i++)B[i]=A[i+1]-A[i];
    B[n-1]=A[0]-A[n-1];
    long long t=(long long)n*(n+1)/2;
    if(s%t){puts("NO");return;}
    s/=t; long long cnt=0;
    for(int i=0;i<n;i++){
        if((s-B[i])%n||s-B[i]<0){puts("NO");return;}
        cnt+=(s-B[i])/n;
    }if(cnt!=s){puts("NO");return;}
    puts("YES");
}
int main(){
    while(~scanf("%d",&n))solve();
    return 0;
}