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;
const int maxn = 2e5 + 10;
int n, m;
vector<int> G[maxn];
bool check(int x) {
  for (int i = 0; i < n; ++i) {
    if (G[i] != G[(i + x) % n]) {
      return false;
    }
  }
  return true;
}
bool solve() {
  for (int i = 1; i < n; ++i) {
    if (n % i == 0) {
      if (check(i)) {
        return true;
      }
    }
  }
  return false;
}
int main() {
  while (~scanf("%d %d", &n, &m)) {
    for (int i = 0; i < n; ++i) {
      G[i].clear();
    }
    for (int i = 1, x, y; i <= m; ++i) {
      scanf("%d %d", &x, &y);
      if (x > y) swap(x, y);
      G[x - 1].push_back(y - x);
      G[y - 1].push_back(x + n - y);
    }
    for (int i = 0; i < n; ++i) {
      sort(G[i].begin(), G[i].end());
    }
    puts(solve() ? "Yes" : "No");
  }
  return 0;
}
