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<int> S[100010];
int n, m;
pair<int, int> v[200020];
bool ok(int k) {
  if (k == n) return false;
  for (int t = 0; t < m; t++) {
    int i = v[t].first, j = v[t].second;
    if (S[(i + k) % n].find((j + k) % n) == S[(i + k) % n].end()) return false;
  }
  return true;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    S[a].insert(b);
    S[b].insert(a);
    v[i] = pair<int, int>(a, b);
  }
  bool da = false;
  for (int i = 1; i <= n / i; i++)
    if (n % i == 0) {
      if (ok(i))
        da = true;
      else if (n / i != i and ok(n / i))
        da = true;
      if (da) break;
    }
  printf("%s\n", da ? "Yes" : "No");
  return 0;
}
