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;
long long int exp(long long int a, long long int b, long long int m);
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, m;
  cin >> n >> m;
  long long int a, b;
  long long int a1[m], b1[m];
  map<pair<long long int, long long int>, bool> sb;
  for (long long int i = 0; i < m; i++) {
    cin >> a >> b;
    a1[i] = a;
    b1[i] = b;
    sb[make_pair(a, b)] = 1;
    sb[make_pair(b, a)] = 1;
  }
  long long int sq = sqrt(n);
  bool ans = 0;
  for (long long int j = 1; j <= sq; j++) {
    if (n % j != 0) continue;
    long long int d = j;
    bool ans1 = 1;
    for (long long int i = 0; i < m; i++) {
      long long int f1 = (a1[i] + d - 1) % n + 1;
      long long int f2 = (b1[i] + d - 1) % n + 1;
      if (sb[make_pair(f1, f2)] != 1) ans1 = 0;
      if (ans1 == 0) break;
    }
    if (ans1 == 1) {
      ans = 1;
      break;
    }
    ans1 = 1;
    d = n / j;
    if (d == n) continue;
    for (long long int i = 0; i < m; i++) {
      long long int f1 = (a1[i] + d - 1) % n + 1;
      long long int f2 = (b1[i] + d - 1) % n + 1;
      if (sb[make_pair(f1, f2)] != 1) ans1 = 0;
      if (ans1 == 0) break;
    }
    if (ans1 == 1) {
      ans = 1;
      break;
    }
  }
  if (ans == 1)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}
long long int exp(long long int a, long long int b, long long int m) {
  if (b == 0) {
    return 1;
  }
  long long int temp = exp(a, b / 2, m);
  temp = (temp * temp) % m;
  if (b & 1) {
    return (temp * (a % m)) % m;
  }
  return temp;
}
