#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long N = 1e6 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  set<long long> adj[n];
  while (m-- > 0) {
    long long a, b;
    cin >> a >> b;
    a -= 1, b -= 1;
    adj[a].insert(b);
    adj[b].insert(a);
  }
  for (long long i = 1; i < n; i++) {
    if (n % i) continue;
    bool flg = true;
    for (long long u = 0; u < n; u++) {
      for (set<long long>::iterator itr = adj[u].begin(); itr != adj[u].end();
           itr++) {
        long long v = *itr;
        if (adj[(u + i) % n].find((v + i) % n) == adj[(u + i) % n].end())
          flg = false;
      }
    }
    if (flg) return cout << "Yes\n", 0;
  }
  cout << "No";
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long N = 1e6 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  set<long long> adj[n];
  while (m-- > 0) {
    long long a, b;
    cin >> a >> b;
    a -= 1, b -= 1;
    adj[a].insert(b);
    adj[b].insert(a);
  }
  for (long long i = 1; i < n; i++) {
    if (n % i) continue;
    bool flg = true;
    for (long long u = 0; u < n; u++) {
      for (set<long long>::iterator itr = adj[u].begin(); itr != adj[u].end();
           itr++) {
        long long v = *itr;
        if (adj[(u + i) % n].find((v + i) % n) == adj[(u + i) % n].end())
          flg = false;
      }
    }
    if (flg) return cout << "Yes\n", 0;
  }
  cout << "No";
}

```