#include <cstdio>

const int maxn = 1e5 + 10;

int A[maxn];
int B[maxn];

int main()
{
	int N;
	scanf("%d", &N);
	long long sum = 0;
	for (int i = 0; i < N; ++i)
		scanf("%d", &A[i]), sum += A[i];
	long long t = 1ll * N * (N + 1) / 2;
	if (sum % t)
	{
		puts("NO");
		return 0;
	}
	long long cnt = sum / t;
	for (int i = 0; i < N; ++i)
		B[i] = A[(i + 1) % N] - A[i] - cnt;
	bool flag = true;
	for (int i = 0; i < N; ++i)
		if (B[i] > 0 || (-B[i]) % N)
			flag = false;
	puts(flag ? "YES" : "NO");
	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 <cstdio>

const int maxn = 1e5 + 10;

int A[maxn];
int B[maxn];

int main()
{
	int N;
	scanf("%d", &N);
	long long sum = 0;
	for (int i = 0; i < N; ++i)
		scanf("%d", &A[i]), sum += A[i];
	long long t = 1ll * N * (N + 1) / 2;
	if (sum % t)
	{
		puts("NO");
		return 0;
	}
	long long cnt = sum / t;
	for (int i = 0; i < N; ++i)
		B[i] = A[(i + 1) % N] - A[i] - cnt;
	bool flag = true;
	for (int i = 0; i < N; ++i)
		if (B[i] > 0 || (-B[i]) % N)
			flag = false;
	puts(flag ? "YES" : "NO");
	return 0;
}

```