#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(m), b(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i] >> b[i];
    a[i]--;
    b[i]--;
    if (a[i] > b[i]) {
      swap(a[i], b[i]);
    }
  }
  vector<int> d;
  for (int i = 1; i <= ceil(sqrt(n)); i++) {
    if (n % i == 0) {
      d.push_back(i);
      if (i * i != n) {
        d.push_back(n / i);
      }
    }
  }
  set<pair<int, int>> s;
  for (int i = 0; i < m; i++) {
    s.insert({a[i], b[i]});
  }
  for (auto &k : d) {
    if (k == n) {
      continue;
    }
    bool f = true;
    for (int i = 0; i < m; i++) {
      int A = (a[i] + k) % n, B = (b[i] + k) % n;
      f &= (s.find({min(A, B), max(A, B)}) != s.end());
    }
    if (f) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No\n";
  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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(m), b(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i] >> b[i];
    a[i]--;
    b[i]--;
    if (a[i] > b[i]) {
      swap(a[i], b[i]);
    }
  }
  vector<int> d;
  for (int i = 1; i <= ceil(sqrt(n)); i++) {
    if (n % i == 0) {
      d.push_back(i);
      if (i * i != n) {
        d.push_back(n / i);
      }
    }
  }
  set<pair<int, int>> s;
  for (int i = 0; i < m; i++) {
    s.insert({a[i], b[i]});
  }
  for (auto &k : d) {
    if (k == n) {
      continue;
    }
    bool f = true;
    for (int i = 0; i < m; i++) {
      int A = (a[i] + k) % n, B = (b[i] + k) % n;
      f &= (s.find({min(A, B), max(A, B)}) != s.end());
    }
    if (f) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No\n";
  return 0;
}

```