#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<vector<int>> id(n / 2);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b) swap(a, b);
    if (b - a <= n / 2) {
      id[b - a - 1].push_back(a);
    } else {
      id[n - b + a - 1].push_back(b);
      swap(a, b);
    }
    if (n % 2 == 0 && (b - a + n) % n == n / 2) id[n / 2 - 1].push_back(b);
  }
  int r = 1;
  for (int i = 0; i < n / 2; i++) {
    if (id[i].size() == 1) {
      cout << "No";
      return 0;
    }
    sort(id[i].begin(), id[i].end(),
         [&](const int x, const int y) { return x < y; });
    int j = r;
    for (; j < n; j += r) {
      int k = 0;
      for (; k < id[i].size(); k++) {
        int next = (id[i][k] + j) % n;
        if (!next) next = n;
        if (!binary_search(id[i].begin(), id[i].end(), next)) break;
      }
      if (k == id[i].size()) break;
    }
    if (j < n)
      r = j;
    else {
      cout << "No";
      return 0;
    }
  }
  cout << "Yes";
  return 0;
}

### Prompt

Please formulate a cpp solution to 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<vector<int>> id(n / 2);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b) swap(a, b);
    if (b - a <= n / 2) {
      id[b - a - 1].push_back(a);
    } else {
      id[n - b + a - 1].push_back(b);
      swap(a, b);
    }
    if (n % 2 == 0 && (b - a + n) % n == n / 2) id[n / 2 - 1].push_back(b);
  }
  int r = 1;
  for (int i = 0; i < n / 2; i++) {
    if (id[i].size() == 1) {
      cout << "No";
      return 0;
    }
    sort(id[i].begin(), id[i].end(),
         [&](const int x, const int y) { return x < y; });
    int j = r;
    for (; j < n; j += r) {
      int k = 0;
      for (; k < id[i].size(); k++) {
        int next = (id[i][k] + j) % n;
        if (!next) next = n;
        if (!binary_search(id[i].begin(), id[i].end(), next)) break;
      }
      if (k == id[i].size()) break;
    }
    if (j < n)
      r = j;
    else {
      cout << "No";
      return 0;
    }
  }
  cout << "Yes";
  return 0;
}

```