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;

int main(){
 long long n,i,a[100010],b[100010],sum=0,mx=0;
 cin >> n;
 for(i=0;i<n;i++){
     cin >> a[i];
     sum += a[i];
 }

if(sum%(n*(n+1)/2)!=0){
    cout << "NO" << endl;
    return 0;
}
long long k = sum/(n*(n+1)/2);

 for(i=0;i<n;i++){
     b[i] = a[0]+i*k-a[i];
     if(b[i]%n!=0){
         cout <<"NO" << endl;
         return 0;
     }
 }

for(i=0;i<n-1;i++){
    if(b[i]>b[i+1] ){
        cout << "NO" << endl;
        return 0;
    }
}

if(b[n-1]/n>k){
    cout << "NO" << endl;
    return 0;
}
    
    cout <<"YES" << endl;
}
