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;
struct pair_hash {
  template <class T1, class T2>
  size_t operator()(pair<T1, T2> const &pair) const {
    size_t h1 = std::hash<T1>()(pair.first);
    size_t h2 = std::hash<T2>()(pair.second);
    return h1 ^ h2;
  }
};
unordered_set<pair<int, int>, pair_hash> segmentos;
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  vector<pair<int, int> > a(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i].first >> a[i].second;
    segmentos.insert({a[i].first, a[i].second});
  }
  for (int i = 1; i < n; i++) {
    if (n % i) continue;
    bool solved = true;
    for (int j = 0; j < m && solved; j++) {
      int optA = a[j].first;
      int optB = a[j].second;
      optA = (a[j].first + i > n ? 1 : 0) + (a[j].first + i) % (n + 1);
      optB = (a[j].second + i > n ? 1 : 0) + (a[j].second + i) % (n + 1);
      if (segmentos.find({optA, optB}) == segmentos.end() &&
          segmentos.find({optB, optA}) == segmentos.end()) {
        solved = false;
      }
    }
    if (solved) {
      printf("Yes");
      exit(0);
    }
  }
  printf("No");
  return 0;
}
