#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
  ll n;
  cin>>n;
  ll a[n],b[n];
  for(ll i=0;i<n;i++) cin>>a[i];
  ll sum,mi=n*(n+1)/2;
  for(ll i=0;i<n;i++) sum+=a[i];
  if(sum%mi){
    cout<<"NO"<<endl;
    return 0;
  }
  bool ch=0;
  for(ll i=0;i<n;i++) ch|=!(sum/mi<=a[i]&&a[i]<=sum/mi*n);
  if(ch){
    cout<<"NO"<<endl;
    return 0;
  }
  ll cnt=0;
  bool ff=1;
  for(ll i=0;i<n;i++){
    b[i]=a[(i+1)%n]-a[i];
    if((sum/mi-b[i])%n){
      ff=0;break;
    }
    cnt+=(sum/mi-b[i])/n;
  }
  cout<<(ff&&cnt==sum/mi?"YES":"NO")<<endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
int main(){
  ll n;
  cin>>n;
  ll a[n],b[n];
  for(ll i=0;i<n;i++) cin>>a[i];
  ll sum,mi=n*(n+1)/2;
  for(ll i=0;i<n;i++) sum+=a[i];
  if(sum%mi){
    cout<<"NO"<<endl;
    return 0;
  }
  bool ch=0;
  for(ll i=0;i<n;i++) ch|=!(sum/mi<=a[i]&&a[i]<=sum/mi*n);
  if(ch){
    cout<<"NO"<<endl;
    return 0;
  }
  ll cnt=0;
  bool ff=1;
  for(ll i=0;i<n;i++){
    b[i]=a[(i+1)%n]-a[i];
    if((sum/mi-b[i])%n){
      ff=0;break;
    }
    cnt+=(sum/mi-b[i])/n;
  }
  cout<<(ff&&cnt==sum/mi?"YES":"NO")<<endl;
  return 0;
}

```