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;
set<pair<int, int> > s;
vector<int> k;
void divs(int n) {
  int limit = sqrt(n);
  for (int i = 1; i <= limit; i += 1) {
    if (n % i == 0) {
      k.push_back(i);
      if (n / i != i && n / i != n) k.push_back(n / i);
    }
  }
}
int main() {
  int n, seg;
  scanf("%d%d", &n, &seg);
  for (int i = 0; i < seg; i += 1) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    s.insert(pair<int, int>(a, b));
  }
  divs(n);
  int ki = k.size();
  for (int i = 0; i < ki; i += 1) {
    bool res = true;
    for (auto x : s) {
      int a = x.first + k[i];
      int b = x.second + k[i];
      if (a > n) a -= n;
      if (b > n) b -= n;
      if (a > b) swap(a, b);
      if (s.find(pair<int, int>(a, b)) == s.end()) {
        res = false;
        break;
      }
    }
    if (res) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}
