#include <bits/stdc++.h>
using namespace std;
int N, M;
vector<int> arr[100005];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  for (int i = 1; i <= M; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    if (true) {
      int d = a - b;
      if (d < 0) {
        d += N;
      }
      arr[a].push_back(d);
    }
    if (true) {
      int d = b - a;
      if (d < 0) {
        d += N;
      }
      arr[b].push_back(d);
    }
  }
  for (int i = 0; i < N; i++) {
    sort(arr[i].begin(), arr[i].end());
  }
  for (int d = 1; d < N; d++) {
    if (N % d) {
      continue;
    }
    bool b = 1;
    for (int i = 0, j = d; i < N; i++, j = (j + 1) % N) {
      if (arr[i].size() != arr[j].size()) {
        b = 0;
        break;
      }
      int s = arr[i].size();
      for (int k = 0; k < s; k++) {
        if (arr[i][k] != arr[j][k]) {
          b = 0;
          break;
        }
      }
    }
    if (b) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
}

### 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 N, M;
vector<int> arr[100005];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  for (int i = 1; i <= M; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    if (true) {
      int d = a - b;
      if (d < 0) {
        d += N;
      }
      arr[a].push_back(d);
    }
    if (true) {
      int d = b - a;
      if (d < 0) {
        d += N;
      }
      arr[b].push_back(d);
    }
  }
  for (int i = 0; i < N; i++) {
    sort(arr[i].begin(), arr[i].end());
  }
  for (int d = 1; d < N; d++) {
    if (N % d) {
      continue;
    }
    bool b = 1;
    for (int i = 0, j = d; i < N; i++, j = (j + 1) % N) {
      if (arr[i].size() != arr[j].size()) {
        b = 0;
        break;
      }
      int s = arr[i].size();
      for (int k = 0; k < s; k++) {
        if (arr[i][k] != arr[j][k]) {
          b = 0;
          break;
        }
      }
    }
    if (b) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
}

```