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;
set<pair<int, int> > segs;
vector<int> listaDivisores;
int main() {
  int n, m;
  cin >> n >> m;
  int a, b;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    segs.insert(make_pair(a, b));
  }
  for (int i = 1; i * i <= n; i++)
    if (n % i == 0) {
      listaDivisores.push_back(i);
      if (i * i != n && i != 1) listaDivisores.push_back(n / i);
    }
  bool res;
  for (auto x : listaDivisores) {
    res = true;
    for (auto s : segs) {
      int ini = s.first + x;
      int end = s.second + x;
      if (ini > n) ini = (ini % (n + 1)) + 1;
      if (end > n) end = (end % (n + 1)) + 1;
      if (segs.find(make_pair(ini, end)) == segs.end() &&
          segs.find(make_pair(end, ini)) == segs.end())
        res = false;
      if (!res) break;
    }
    if (res) break;
  }
  if (res)
    cout << "Yes\n";
  else
    cout << "No\n";
  return 0;
}
