#include <bits/stdc++.h>
using namespace std;

long long N;
#define MAXN 100000
long long A[MAXN];

int main() {
  scanf("%lld", &N);
  long long sum = 0;
  for(int i = 0; i < N; ++i) {
    scanf("%lld", &A[i]);
    sum += A[i];
  }
  if(sum % (N * (N + 1) / 2) != 0) { printf("NO\n"); return 0;}
  long long k = sum / (N * (N + 1) / 2);
  long long sum_d = 0;
  for(int i = 0; i < N; ++i) {
    long long d = A[i] + k - A[(i + 1) % N];
    if(d < 0 || d % N != 0) { printf("NO\n"); return 0;}
    sum_d += d / N;
  }
  printf("%s\n", sum_d == k ? "YES" : "NO");
  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;

long long N;
#define MAXN 100000
long long A[MAXN];

int main() {
  scanf("%lld", &N);
  long long sum = 0;
  for(int i = 0; i < N; ++i) {
    scanf("%lld", &A[i]);
    sum += A[i];
  }
  if(sum % (N * (N + 1) / 2) != 0) { printf("NO\n"); return 0;}
  long long k = sum / (N * (N + 1) / 2);
  long long sum_d = 0;
  for(int i = 0; i < N; ++i) {
    long long d = A[i] + k - A[(i + 1) % N];
    if(d < 0 || d % N != 0) { printf("NO\n"); return 0;}
    sum_d += d / N;
  }
  printf("%s\n", sum_d == k ? "YES" : "NO");
  return 0;
}

```