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 n, m, u, v;
vector<pair<int, int>> a;
bool ch(int c) {
  vector<pair<int, int>> b = a;
  for (auto &i : b) {
    i.first += c;
    i.second += c;
    if (i.first > n) i.first -= n;
    if (i.second > n) i.second -= n;
    if (i.first > i.second) swap(i.first, i.second);
  }
  sort(b.begin(), b.end());
  return b == a;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u, &v);
    if (u > v) swap(u, v);
    a.push_back(make_pair(u, v));
  }
  sort(a.begin(), a.end());
  int x = sqrt(n);
  if (ch(1)) return printf("Yes"), 0;
  for (int i = x; i >= 2; i--) {
    if (n % i == 0) {
      if (ch(i) || ch(n / i)) {
        printf("Yes");
        return 0;
      }
    }
  }
  printf("No");
  return 0;
}
