#include <bits/stdc++.h>
using namespace std;
vector<int> edge[100010];
int len1[100010], len2[100010];
int n, m;
int len(int a, int b) {
  if (a > b)
    return min(a - b, b - a + n);
  else
    return min(b - a, a - b + n);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  for (int i = 1; i <= n / 2; i++) {
    if (n % i != 0) continue;
    int flag = 1;
    for (int j = 1; j <= n; j++) {
      if (edge[j].size() != edge[(j + i - 1) % n + 1].size()) {
        flag = 0;
        break;
      }
      for (int k = 0; k < edge[j].size(); k++) {
        len1[k] = len(edge[j][k], j);
        len2[k] = len(edge[(j + i - 1) % n + 1][k], (j + i - 1) % n + 1);
      }
      sort(len1, len1 + edge[j].size());
      sort(len2, len2 + edge[j].size());
      for (int k = 0; k < edge[j].size(); k++) {
        if (len1[k] != len2[k]) {
          flag = 0;
          break;
        }
      }
    }
    if (flag) {
      printf("Yes");
      return 0;
    }
  }
  printf("No");
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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;
vector<int> edge[100010];
int len1[100010], len2[100010];
int n, m;
int len(int a, int b) {
  if (a > b)
    return min(a - b, b - a + n);
  else
    return min(b - a, a - b + n);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  for (int i = 1; i <= n / 2; i++) {
    if (n % i != 0) continue;
    int flag = 1;
    for (int j = 1; j <= n; j++) {
      if (edge[j].size() != edge[(j + i - 1) % n + 1].size()) {
        flag = 0;
        break;
      }
      for (int k = 0; k < edge[j].size(); k++) {
        len1[k] = len(edge[j][k], j);
        len2[k] = len(edge[(j + i - 1) % n + 1][k], (j + i - 1) % n + 1);
      }
      sort(len1, len1 + edge[j].size());
      sort(len2, len2 + edge[j].size());
      for (int k = 0; k < edge[j].size(); k++) {
        if (len1[k] != len2[k]) {
          flag = 0;
          break;
        }
      }
    }
    if (flag) {
      printf("Yes");
      return 0;
    }
  }
  printf("No");
  return 0;
}

```