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>
int mod, n, x, y, good;
std::vector<int> f;
std::set<std::pair<int, int> > set;
void factorize() {
  for (int i = 1; i < mod; i++)
    if (mod % i == 0) f.push_back(i);
}
std::pair<int, int> make_pair(int i, int j) {
  return {std::min(i, j), std::max(i, j)};
}
int main() {
  scanf("%d%d", &mod, &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x, &y);
    set.insert(make_pair(x - 1, y - 1));
  }
  factorize();
  for (int k : f) {
    good = true;
    for (auto p : set)
      if (!set.count(make_pair((p.first + k) % mod, (p.second + k) % mod))) {
        good = false;
        break;
      }
    if (good) break;
  }
  printf("%s\n", good ? "Yes" : "No");
  return 0;
}
