#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;

int main()
{
ll n ;
cin >> n ;
ll sum=0;
vector<ll> a(n);
rep(i, n){
  cin >> a[i];
  sum+=a[i];
}

if(sum%((1+n)*n/2)!=0){cout << "NO" << endl;return 0;}
sum/=(1+n)*n/2;
rep(i,n-1){
  if((a[i]+sum-a[i+1])%n!=0){cout << "NO" << endl;return 0;}
  if(a[i+1]-a[i]>sum){cout << "NO" << endl;return 0;}
}
if((a[n-1]+sum-a[0])%n!=0){cout << "NO" << endl;return 0;}
if(a[n-1]-a[0]>sum){cout << "NO" << endl;return 0;}
cout << "YES" << endl;
return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;

int main()
{
ll n ;
cin >> n ;
ll sum=0;
vector<ll> a(n);
rep(i, n){
  cin >> a[i];
  sum+=a[i];
}

if(sum%((1+n)*n/2)!=0){cout << "NO" << endl;return 0;}
sum/=(1+n)*n/2;
rep(i,n-1){
  if((a[i]+sum-a[i+1])%n!=0){cout << "NO" << endl;return 0;}
  if(a[i+1]-a[i]>sum){cout << "NO" << endl;return 0;}
}
if((a[n-1]+sum-a[0])%n!=0){cout << "NO" << endl;return 0;}
if(a[n-1]-a[0]>sum){cout << "NO" << endl;return 0;}
cout << "YES" << endl;
return 0;
}

```