#include <bits/stdc++.h>
const int mod = 1000000000 + 7;
using namespace std;
void no(int p = -1) {
  cout << "NIE"
       << "\n";
}
int n, m, u, v;
int gcd(int a, int b) { return (b ? gcd(b, a % b) : a); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  vector<pair<int, int>> vec;
  for (int i = 0; i < int(m); i++) {
    cin >> u >> v;
    u--;
    v--;
    if (u > v) swap(u, v);
    vec.emplace_back(u, v);
  }
  sort(vec.begin(), vec.end());
  int t = n;
  for (int i = 2; i <= t; i++) {
    if (t % i == 0) {
      vector<pair<int, int>> v2;
      for (auto x : vec) {
        x.first = (x.first + n / i) % n;
        x.second = (x.second + n / i) % n;
        if (x.first > x.second) swap(x.first, x.second);
        v2.emplace_back(x.first, x.second);
      }
      sort(v2.begin(), v2.end());
      if (v2 == vec) {
        cout << "Yes";
        exit(0);
      }
      while (t % i == 0) t /= i;
    }
  }
  cout << "No";
  return 0;
}

### Prompt

Please create a solution in CPP 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>
const int mod = 1000000000 + 7;
using namespace std;
void no(int p = -1) {
  cout << "NIE"
       << "\n";
}
int n, m, u, v;
int gcd(int a, int b) { return (b ? gcd(b, a % b) : a); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  vector<pair<int, int>> vec;
  for (int i = 0; i < int(m); i++) {
    cin >> u >> v;
    u--;
    v--;
    if (u > v) swap(u, v);
    vec.emplace_back(u, v);
  }
  sort(vec.begin(), vec.end());
  int t = n;
  for (int i = 2; i <= t; i++) {
    if (t % i == 0) {
      vector<pair<int, int>> v2;
      for (auto x : vec) {
        x.first = (x.first + n / i) % n;
        x.second = (x.second + n / i) % n;
        if (x.first > x.second) swap(x.first, x.second);
        v2.emplace_back(x.first, x.second);
      }
      sort(v2.begin(), v2.end());
      if (v2 == vec) {
        cout << "Yes";
        exit(0);
      }
      while (t % i == 0) t /= i;
    }
  }
  cout << "No";
  return 0;
}

```