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() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, a, b;
  cin >> n >> m;
  map<pair<int, int>, int> mp;
  while (m--) {
    cin >> a >> b;
    a--;
    b--;
    mp[{min(a, b), max(a, b)}] = 1;
  }
  int cnt = 0, f;
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      cnt++;
      f = 0;
      for (auto j : mp) {
        int one = min((j.first.first + i) % n, (j.first.second + i) % n);
        int two = max((j.first.first + i) % n, (j.first.second + i) % n);
        if (mp.find({one, two}) == mp.end()) {
          f = 1;
          break;
        }
        mp[{one, two}]++;
      }
      if (f == 0) {
        cout << "Yes";
        return 0;
      }
      cnt++;
      if (i == 1) continue;
      f = 0;
      for (auto j : mp) {
        int one =
            min((j.first.first + n / i) % n, (j.first.second + n / i) % n);
        int two =
            max((j.first.first + n / i) % n, (j.first.second + n / i) % n);
        if (mp.find({one, two}) == mp.end()) {
          f = 1;
          break;
        }
        mp[{one, two}]++;
      }
      if (f == 0) {
        cout << "Yes";
        return 0;
      }
    }
  }
  cout << "No";
  return 0;
}
