#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
long long a[N];
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > v;
  int x, y;
  set<long long> s;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    x--, y--;
    v.push_back({x, y});
    s.insert(x * N + y);
    s.insert(y * N + x);
  }
  int yo = sqrtl(n);
  for (int i = 1; i <= yo; i++) {
    if (n % i == 0) {
      int flag = 0;
      for (auto j : v) {
        x = j.first, y = j.second;
        x = (x + i) % n, y = (y + i) % n;
        if (s.find(x * N + y) == s.end() && s.find(y * N + x) == s.end()) {
          flag = 1;
          break;
        }
      }
      if (!flag) {
        return cout << "Yes", 0;
      }
    }
    int temp = i;
    if (n % (n / i) == 0 && i != 1) {
      i = n / i;
      int flag = 0;
      for (auto j : v) {
        x = j.first, y = j.second;
        x = (x + i) % n, y = (y + i) % n;
        if (s.find(x * N + y) == s.end() && s.find(y * N + x) == s.end()) {
          flag = 1;
          break;
        }
      }
      if (!flag) {
        return cout << "Yes", 0;
      }
    }
    i = temp;
  }
  cout << "No";
  return 0;
}

### Prompt

Please formulate 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;
const int N = 3e5 + 5;
long long a[N];
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > v;
  int x, y;
  set<long long> s;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    x--, y--;
    v.push_back({x, y});
    s.insert(x * N + y);
    s.insert(y * N + x);
  }
  int yo = sqrtl(n);
  for (int i = 1; i <= yo; i++) {
    if (n % i == 0) {
      int flag = 0;
      for (auto j : v) {
        x = j.first, y = j.second;
        x = (x + i) % n, y = (y + i) % n;
        if (s.find(x * N + y) == s.end() && s.find(y * N + x) == s.end()) {
          flag = 1;
          break;
        }
      }
      if (!flag) {
        return cout << "Yes", 0;
      }
    }
    int temp = i;
    if (n % (n / i) == 0 && i != 1) {
      i = n / i;
      int flag = 0;
      for (auto j : v) {
        x = j.first, y = j.second;
        x = (x + i) % n, y = (y + i) % n;
        if (s.find(x * N + y) == s.end() && s.find(y * N + x) == s.end()) {
          flag = 1;
          break;
        }
      }
      if (!flag) {
        return cout << "Yes", 0;
      }
    }
    i = temp;
  }
  cout << "No";
  return 0;
}

```