#include<stdio.h>

typedef long long lld;
lld N, NN, ba[101010], sum;

int main(){
	scanf("%lld", &N), NN=N*(N+1)/2;
	for(int i=0; i<N; i++)scanf("%lld", &ba[i]), sum+=ba[i];
	if(sum % NN)puts("NO");
	else{
		sum /= NN;
		for(int i=0; i<N; i++){
			lld t = ba[i]+sum-ba[(i+1)%N];
			if(t<0 || t%N){puts("NO"); return 0;}
		}
		puts("YES");
	}
	return 0;
}
### Prompt

Generate 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<stdio.h>

typedef long long lld;
lld N, NN, ba[101010], sum;

int main(){
	scanf("%lld", &N), NN=N*(N+1)/2;
	for(int i=0; i<N; i++)scanf("%lld", &ba[i]), sum+=ba[i];
	if(sum % NN)puts("NO");
	else{
		sum /= NN;
		for(int i=0; i<N; i++){
			lld t = ba[i]+sum-ba[(i+1)%N];
			if(t<0 || t%N){puts("NO"); return 0;}
		}
		puts("YES");
	}
	return 0;
}
```