#include <iostream>
#include <cstdio>
#define N 100005
using namespace std;

typedef long long ll;
ll n, s, c, a[N], b[N];
int main()
{
	ll i;
	cin >> n;
	for (i = 1; i <= n; i++) scanf("%lld", &a[i]), s += a[i], b[i] = a[i] - a[i - 1];
	b[1] = a[1] - a[n];
	if (s % (n * (n + 1) / 2)) {cout << "NO"; return 0;}
	s /= n * (n + 1) / 2;
	for (i = 1; i <= n; i++) {
		if (s < b[i] || (s - b[i]) % n) {cout << "NO"; return 0;}
//		c += (s - b[i]) / n;
	}
//	if (c != s) {cout << "NO"; return 0;}
    cout << "YES" << endl;
    return 0;
}
### Prompt

In CPP, your task is to solve 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 <iostream>
#include <cstdio>
#define N 100005
using namespace std;

typedef long long ll;
ll n, s, c, a[N], b[N];
int main()
{
	ll i;
	cin >> n;
	for (i = 1; i <= n; i++) scanf("%lld", &a[i]), s += a[i], b[i] = a[i] - a[i - 1];
	b[1] = a[1] - a[n];
	if (s % (n * (n + 1) / 2)) {cout << "NO"; return 0;}
	s /= n * (n + 1) / 2;
	for (i = 1; i <= n; i++) {
		if (s < b[i] || (s - b[i]) % n) {cout << "NO"; return 0;}
//		c += (s - b[i]) / n;
	}
//	if (c != s) {cout << "NO"; return 0;}
    cout << "YES" << endl;
    return 0;
}
```