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;
const int nmax=100005;
long long sum,a[nmax],d[nmax];
int n,i;
int main()
{
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>a[i];
        sum+=a[i];
    }
    if(sum%(1LL*n*(n+1)/2))
    {
        cout<<"NO";
        return 0;
    }
    long long k=(sum/((1LL*n*(n+1)/2)));
    a[n+1]=a[1];
    for(i=1;i<=n;i++)
    {
        d[i]=a[i+1]-a[i];
        if(k-d[i]<0||(k-d[i])%n)
        {
            cout<<"NO";
            return 0;
        }
    }
    cout<<"YES";
    return 0;
}
