#include <bits/stdc++.h>
using namespace std;
using std::cin;
using std::cout;
long long MOD = 998244353;
int main() {
  int n, m;
  cin >> n >> m;
  pair<int, int> arr[m];
  set<pair<int, int> > s;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a %= n;
    b %= n;
    if (b < a) swap(a, b);
    arr[i] = {a, b};
    s.insert(arr[i]);
  }
  for (int i = 1; i <= sqrt(n) + 1; i++) {
    if (n % i == 0) {
      bool a = true;
      for (int c = 0; c < m; c++) {
        int firstelem = (arr[c].first + i) % n;
        int secondelem = (arr[c].second + i) % n;
        if (firstelem > secondelem) swap(firstelem, secondelem);
        if (s.find({firstelem, secondelem}) == s.end()) {
          a = false;
          break;
        }
      }
      if (a) {
        cout << "Yes"
             << "\n";
        return 0;
      }
      if (i != 1) {
        a = true;
        for (int c = 0; c < m; c++) {
          int firstelem = (arr[c].first + n / i) % n;
          int secondelem = (arr[c].second + n / i) % n;
          if (firstelem > secondelem) swap(firstelem, secondelem);
          if (s.find({firstelem, secondelem}) == s.end()) {
            a = false;
            break;
          }
        }
      }
      if (a) {
        cout << "Yes"
             << "\n";
        return 0;
      }
    }
  }
  cout << "No"
       << "\n";
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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;
using std::cin;
using std::cout;
long long MOD = 998244353;
int main() {
  int n, m;
  cin >> n >> m;
  pair<int, int> arr[m];
  set<pair<int, int> > s;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a %= n;
    b %= n;
    if (b < a) swap(a, b);
    arr[i] = {a, b};
    s.insert(arr[i]);
  }
  for (int i = 1; i <= sqrt(n) + 1; i++) {
    if (n % i == 0) {
      bool a = true;
      for (int c = 0; c < m; c++) {
        int firstelem = (arr[c].first + i) % n;
        int secondelem = (arr[c].second + i) % n;
        if (firstelem > secondelem) swap(firstelem, secondelem);
        if (s.find({firstelem, secondelem}) == s.end()) {
          a = false;
          break;
        }
      }
      if (a) {
        cout << "Yes"
             << "\n";
        return 0;
      }
      if (i != 1) {
        a = true;
        for (int c = 0; c < m; c++) {
          int firstelem = (arr[c].first + n / i) % n;
          int secondelem = (arr[c].second + n / i) % n;
          if (firstelem > secondelem) swap(firstelem, secondelem);
          if (s.find({firstelem, secondelem}) == s.end()) {
            a = false;
            break;
          }
        }
      }
      if (a) {
        cout << "Yes"
             << "\n";
        return 0;
      }
    }
  }
  cout << "No"
       << "\n";
  return 0;
}

```