#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, m;
  cin >> n >> m;
  vector<set<long long>> v((size_t)n);
  auto add = [&](long long a, long long b) { v[a].insert((b - a + n) % n); };
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    a--, b--;
    add(a, b);
    add(b, a);
  }
  bool o = false;
  for (long long j = 2, k = n; j <= k; j++) {
    if (k % j == 0) {
      bool g = true;
      for (long long i = 0; i < n - n / j; i++) {
        if (v[i] != v[i + n / j]) {
          g = false;
          break;
        }
      }
      if (g) {
        o = true;
        break;
      }
    }
    while (k % j == 0) {
      k /= j;
    }
  }
  cout << (o ? "Yes" : "No") << endl;
  return 0;
}

### 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, m;
  cin >> n >> m;
  vector<set<long long>> v((size_t)n);
  auto add = [&](long long a, long long b) { v[a].insert((b - a + n) % n); };
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    a--, b--;
    add(a, b);
    add(b, a);
  }
  bool o = false;
  for (long long j = 2, k = n; j <= k; j++) {
    if (k % j == 0) {
      bool g = true;
      for (long long i = 0; i < n - n / j; i++) {
        if (v[i] != v[i + n / j]) {
          g = false;
          break;
        }
      }
      if (g) {
        o = true;
        break;
      }
    }
    while (k % j == 0) {
      k /= j;
    }
  }
  cout << (o ? "Yes" : "No") << endl;
  return 0;
}

```