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 long long mod = 1e9 + 7;
const int MAX = 100005;
long long modexp(long long x, long long n) {
  if (n == 0) return 1LL;
  if (n % 2 == 0) {
    long long y = modexp(x, n / 2) % mod;
    return (y * y) % mod;
  }
  return (x * modexp(x, n - 1) % mod) % mod;
}
long long powr(long long x, long long n) {
  long long ans = 1;
  for (int i = 1; i <= n; i++) ans = ans * x;
  return ans;
}
int main() {
  int n, m;
  scanf("%d", &n), scanf("%d", &m);
  vector<pair<int, int> > v;
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d", &x), scanf("%d", &y);
    x--;
    y--;
    v.push_back(make_pair(min(x, y), max(x, y)));
  }
  vector<int> ans;
  sort(v.begin(), v.end());
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      ans.push_back(i);
    }
  }
  for (int j = 0; j < ans.size(); j++) {
    int flag = 0;
    vector<pair<int, int> > aux;
    int k = ans[j];
    for (int i = 0; i < v.size(); i++) {
      int x = (v[i].first + k) % n;
      int y = (v[i].second + k) % n;
      aux.push_back(make_pair(min(x, y), max(x, y)));
    }
    sort(aux.begin(), aux.end());
    for (int i = 0; i < v.size(); i++) {
      if (aux[i].first == v[i].first && aux[i].second == v[i].second) continue;
      flag = 1;
      break;
    }
    if (flag == 0) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}
