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() {
  set<pair<int, int>> exp, v;
  int mod, n;
  cin >> mod >> n;
  pair<int, int> t;
  for (int i = 0; i < n; i++) {
    cin >> t.first >> t.second;
    if (t.first > t.second) {
      swap(t.first, t.second);
    }
    t.first--;
    t.second--;
    exp.insert(t);
  }
  for (int i = 1; i < mod; i++) {
    if (mod % i == 0) {
      for (auto it : exp) {
        t = it;
        t.first = (t.first + i) % mod;
        t.second = (t.second + i) % mod;
        if (t.first > t.second) {
          swap(t.first, t.second);
        }
        v.insert(t);
      }
      int y = 1;
      for (auto it = exp.begin(), ij = v.begin();
           it != exp.end() && ij != v.end(); it++, ij++) {
        if (*it != *ij) {
          y = 0;
          break;
        }
      }
      if (y) {
        cout << "Yes" << endl;
        return 0;
      }
      v.clear();
    }
  }
  cout << "No" << endl;
  return 0;
}
