#include<bits/stdc++.h>

using namespace std;

int N, a[100009];
long long s;

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d", &N);
for (int i=0; i<N; i++)
    scanf ("%d", &a[i]), s += a[i];
long long u = 1LL * N * (N + 1) / 2;
if (s % u != 0 || s / u > 1e9)
{
    printf ("NO\n");
    return 0;
}
int steps = s / u;
for (int i=0; i<N; i++)
{
    int curr = a[(i + 1) % N] - a[i] - steps;
    if (curr % N != 0 || curr > 0)
    {
        printf ("NO\n");
        return 0;
    }
}
printf ("YES\n");
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<bits/stdc++.h>

using namespace std;

int N, a[100009];
long long s;

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d", &N);
for (int i=0; i<N; i++)
    scanf ("%d", &a[i]), s += a[i];
long long u = 1LL * N * (N + 1) / 2;
if (s % u != 0 || s / u > 1e9)
{
    printf ("NO\n");
    return 0;
}
int steps = s / u;
for (int i=0; i<N; i++)
{
    int curr = a[(i + 1) % N] - a[i] - steps;
    if (curr % N != 0 || curr > 0)
    {
        printf ("NO\n");
        return 0;
    }
}
printf ("YES\n");
return 0;
}

```