Andrewid the Android is a galaxy-famous detective. He is now investigating a case of frauds who make fake copies of the famous Stolp's gears, puzzles that are as famous as the Rubik's cube once was.

Its most important components are a button and a line of n similar gears. Each gear has n teeth containing all numbers from 0 to n - 1 in the counter-clockwise order. When you push a button, the first gear rotates clockwise, then the second gear rotates counter-clockwise, the the third gear rotates clockwise an so on.

Besides, each gear has exactly one active tooth. When a gear turns, a new active tooth is the one following after the current active tooth according to the direction of the rotation. For example, if n = 5, and the active tooth is the one containing number 0, then clockwise rotation makes the tooth with number 1 active, or the counter-clockwise rotating makes the tooth number 4 active.

Andrewid remembers that the real puzzle has the following property: you can push the button multiple times in such a way that in the end the numbers on the active teeth of the gears from first to last form sequence 0, 1, 2, ..., n - 1. Write a program that determines whether the given puzzle is real or fake.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of gears.

The second line contains n digits a1, a2, ..., an (0 ≤ ai ≤ n - 1) — the sequence of active teeth: the active tooth of the i-th gear contains number ai.

Output

In a single line print "Yes" (without the quotes), if the given Stolp's gears puzzle is real, and "No" (without the quotes) otherwise.

Examples

Input

3
1 0 0


Output

Yes


Input

5
4 2 1 4 3


Output

Yes


Input

4
0 2 3 1


Output

No

Note

In the first sample test when you push the button for the first time, the sequence of active teeth will be 2 2 1, when you push it for the second time, you get 0 1 2.
#include <bits/stdc++.h>
int main() {
  int n, a[1005], i, j, count;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  if (a[0] == 0)
    count = 0;
  else
    count = n - a[0];
  for (i = 0; i < count; i++) {
    for (j = 0; j < n; j++) {
      if (j % 2 == 0) {
        if (a[j] == n - 1)
          a[j] = 0;
        else
          a[j] += 1;
      } else {
        if (a[j] == 0)
          a[j] = n - 1;
        else
          a[j] -= 1;
      }
    }
  }
  for (i = 0; i < n; i++) {
    if (a[i] != i) {
      printf("No\n");
      return 0;
    }
  }
  printf("Yes\n");
  return 0;
}
