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>
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<int> > adj(n, vector<int>());
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 0; i < n; i++) {
    sort((adj[i]).begin(), (adj[i]).end());
  }
  set<int> div;
  div.insert(1);
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      div.insert(i);
      div.insert(n / i);
    }
  }
  vector<int> divv = vector<int>((div).begin(), (div).end());
  for (auto it : divv) {
    bool b = true;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < adj[i].size(); j++) {
        int first = (i + it) % n;
        int second = (adj[i][j] + it) % n;
        if (!binary_search((adj[first]).begin(), (adj[first]).end(), second)) {
          b = false;
          break;
        }
      }
    }
    if (b) {
      cout << "Yes" << (char)0x0A;
      return 0;
    }
  }
  cout << "No" << (char)0x0A;
  return 0;
}
