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(NULL);
  cout.tie(NULL);
  long long T = 1;
  while (T--) {
    long long n, m;
    cin >> n >> m;
    vector<pair<long long, long long> > v;
    long long x, y;
    map<pair<long long, long long>, long long> is;
    for (long long i = 0; i < m; i++) {
      cin >> x >> y;
      v.push_back(make_pair(x, y));
      is[make_pair(x, y)] = 1;
    }
    long long f = 0;
    for (long long i = 1; i <= n; i++) {
      if (i == n) continue;
      if (n % i == 0) {
        for (long long j = 0; j < m; j++) {
          long long x = v[j].first;
          long long y = v[j].second;
          x = (x + i) % n;
          y = (y + i) % n;
          if (x == 0) x = n;
          if (y == 0) y = n;
          if (is[make_pair(x, y)] == 0 && is[make_pair(y, x)] == 0) {
            break;
          }
          if (j == m - 1) {
            f = 1;
            break;
          }
        }
        if (f == 1) break;
      }
    }
    if (f == 1)
      cout << "Yes";
    else
      cout << "No";
  }
}
