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;
vector<long long> adj[100005];
long long n;
bool check(long long k) {
  vector<long long> te[100005];
  for (long long i = 1; i <= n; i++) {
    long long x = i + k;
    if (x > n) x -= n;
    for (auto j : adj[i]) {
      long long y = j + k;
      if (y > n) y -= n;
      long long tx = min(x, y), ty = max(x, y);
      te[tx].push_back(ty);
    }
  }
  for (long long i = 1; i <= n; i++) {
    sort(te[i].begin(), te[i].end());
    if (te[i] != adj[i]) return 0;
  }
  return 1;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long m;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    if (x > y) swap(x, y);
    adj[x].push_back(y);
  }
  for (long long i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      if (check(i) || (i != 1 && check(n / i))) {
        cout << "Yes";
        exit(0);
      }
    }
  }
  cout << "No";
}
