#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, a[200005], b[200005], akar, k, x, y;
  bool valid;
  map<pair<int, int>, bool> ada;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &a[i], &b[i]);
    a[i]--;
    b[i]--;
    if (a[i] > b[i]) swap(a[i], b[i]);
    ada[make_pair(a[i], b[i])] = true;
  }
  akar = sqrt(n);
  for (int i = 1; i <= akar; i++)
    if (n % i == 0) {
      if (2 * i > n) break;
      k = i;
      valid = true;
      for (int j = 0; j < m; j++) {
        x = (a[j] + k) % n;
        y = (b[j] + k) % n;
        if (x > y) swap(x, y);
        if (ada.count(make_pair(x, y)) < 1) {
          valid = false;
          break;
        }
      }
      if (valid) {
        puts("Yes");
        return 0;
      }
      k = n / i;
      if ((k == i) || (2 * k > n)) continue;
      valid = true;
      for (int j = 0; j < m; j++) {
        x = (a[j] + k) % n;
        y = (b[j] + k) % n;
        if (x > y) swap(x, y);
        if (ada.count(make_pair(x, y)) < 1) {
          valid = false;
          break;
        }
      }
      if (valid) {
        puts("Yes");
        return 0;
      }
    }
  puts("No");
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

Inaka has a disc, the circumference of which is n units. The circumference is equally divided by n points numbered clockwise from 1 to n, such that points i and i + 1 (1 ≤ i < n) are adjacent, and so are points n and 1.

There are m straight segments on the disc, the endpoints of which are all among the aforementioned n points.

Inaka wants to know if her image is rotationally symmetrical, i.e. if there is an integer k (1 ≤ k < n), such that if all segments are rotated clockwise around the center of the circle by k units, the new image will be the same as the original one.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 200 000) — the number of points and the number of segments, respectively.

The i-th of the following m lines contains two space-separated integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) that describe a segment connecting points a_i and b_i.

It is guaranteed that no segments coincide.

Output

Output one line — "Yes" if the image is rotationally symmetrical, and "No" otherwise (both excluding quotation marks).

You can output each letter in any case (upper or lower).

Examples

Input


12 6
1 3
3 7
5 7
7 11
9 11
11 3


Output


Yes


Input


9 6
4 5
5 6
7 8
8 9
1 2
2 3


Output


Yes


Input


10 3
1 2
3 2
7 2


Output


No


Input


10 2
1 6
2 7


Output


Yes

Note

The first two examples are illustrated below. Both images become the same as their respective original ones after a clockwise rotation of 120 degrees around the center.

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, a[200005], b[200005], akar, k, x, y;
  bool valid;
  map<pair<int, int>, bool> ada;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &a[i], &b[i]);
    a[i]--;
    b[i]--;
    if (a[i] > b[i]) swap(a[i], b[i]);
    ada[make_pair(a[i], b[i])] = true;
  }
  akar = sqrt(n);
  for (int i = 1; i <= akar; i++)
    if (n % i == 0) {
      if (2 * i > n) break;
      k = i;
      valid = true;
      for (int j = 0; j < m; j++) {
        x = (a[j] + k) % n;
        y = (b[j] + k) % n;
        if (x > y) swap(x, y);
        if (ada.count(make_pair(x, y)) < 1) {
          valid = false;
          break;
        }
      }
      if (valid) {
        puts("Yes");
        return 0;
      }
      k = n / i;
      if ((k == i) || (2 * k > n)) continue;
      valid = true;
      for (int j = 0; j < m; j++) {
        x = (a[j] + k) % n;
        y = (b[j] + k) % n;
        if (x > y) swap(x, y);
        if (ada.count(make_pair(x, y)) < 1) {
          valid = false;
          break;
        }
      }
      if (valid) {
        puts("Yes");
        return 0;
      }
    }
  puts("No");
  return 0;
}

```