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

const int maxn = 100010;

typedef long long ll;

int n;
ll a[maxn];

int main() {
	scanf("%d", &n);
	for(int i = 0; i < n; ++i) scanf("%lld", &a[i]);
	ll s1 = accumulate(a, a + n, 0ll);
	ll s0 = 1ll * n * (n + 1) / 2;
	if(s1 % s0) { puts("NO"); return 0; }
	ll m = s1 / s0;
	a[n] = a[0];
	for(int i = 0; i < n; ++i) a[i] = a[i + 1] - a[i] - m;
	for(int i = 0; i < n; ++i) if(a[i] > 0 || a[i] % n) { puts("NO"); return 0; }
	ll s = 0;
	for(int i = 0; i < n; ++i) s += -a[i] / n;
	if(s != m) { puts("NO"); return 0; }
	puts("YES");
	return 0;
}

### Prompt

Create a solution in CPP for 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 <bits/stdc++.h>
using namespace std;

const int maxn = 100010;

typedef long long ll;

int n;
ll a[maxn];

int main() {
	scanf("%d", &n);
	for(int i = 0; i < n; ++i) scanf("%lld", &a[i]);
	ll s1 = accumulate(a, a + n, 0ll);
	ll s0 = 1ll * n * (n + 1) / 2;
	if(s1 % s0) { puts("NO"); return 0; }
	ll m = s1 / s0;
	a[n] = a[0];
	for(int i = 0; i < n; ++i) a[i] = a[i + 1] - a[i] - m;
	for(int i = 0; i < n; ++i) if(a[i] > 0 || a[i] % n) { puts("NO"); return 0; }
	ll s = 0;
	for(int i = 0; i < n; ++i) s += -a[i] / n;
	if(s != m) { puts("NO"); return 0; }
	puts("YES");
	return 0;
}

```