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;
vector<pair<int, int>> segs;
vector<pair<int, int>> temp;
int n, k;
bool solve(int x) {
  temp = segs;
  for (int i = 0; i < k; i++) {
    temp[i].first += x;
    temp[i].second += x;
    if (temp[i].first > n) temp[i].first -= n;
    if (temp[i].second > n) temp[i].second -= n;
    if (temp[i].first > temp[i].second) swap(temp[i].first, temp[i].second);
  }
  sort(temp.begin(), temp.end());
  return (temp == segs);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < k; i++) {
    int l, r;
    cin >> l >> r;
    if (l > r) swap(l, r);
    segs.push_back({l, r});
  }
  if (k == 1) {
    cout << "YES";
    return 0;
  }
  sort(segs.begin(), segs.end());
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      if (solve(n / i) || solve(i)) {
        cout << "YES";
        return 0;
      }
    }
  }
  cout << "NO";
  return 0;
}
