#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, M;
  cin >> N >> M;
  vector<int> A(M), B(M);
  for (int i = 0; i < M; i++) cin >> A[i] >> B[i];
  set<pair<int, int>> se;
  for (int i = 0; i < M; i++) {
    A[i]--;
    B[i]--;
    if (A[i] > B[i]) swap(A[i], B[i]);
    se.insert(make_pair(A[i], B[i]));
  }
  vector<int> divs;
  divs.push_back(1);
  for (int i = 2; i * i <= N; i++) {
    if (N % i == 0) {
      if (i == N / i)
        divs.push_back(i);
      else {
        divs.push_back(i);
        divs.push_back(N / i);
      }
    }
  }
  for (int i = 0; i < divs.size(); i++) {
    int d = divs[i];
    bool ng = false;
    for (int j = 0; j < M; j++) {
      int a = (A[j] + d) % N;
      int b = (B[j] + d) % N;
      if (a > b) swap(a, b);
      if (se.find(make_pair(a, b)) == se.end()) {
        ng = true;
        break;
      }
    }
    if (ng == false) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
  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;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, M;
  cin >> N >> M;
  vector<int> A(M), B(M);
  for (int i = 0; i < M; i++) cin >> A[i] >> B[i];
  set<pair<int, int>> se;
  for (int i = 0; i < M; i++) {
    A[i]--;
    B[i]--;
    if (A[i] > B[i]) swap(A[i], B[i]);
    se.insert(make_pair(A[i], B[i]));
  }
  vector<int> divs;
  divs.push_back(1);
  for (int i = 2; i * i <= N; i++) {
    if (N % i == 0) {
      if (i == N / i)
        divs.push_back(i);
      else {
        divs.push_back(i);
        divs.push_back(N / i);
      }
    }
  }
  for (int i = 0; i < divs.size(); i++) {
    int d = divs[i];
    bool ng = false;
    for (int j = 0; j < M; j++) {
      int a = (A[j] + d) % N;
      int b = (B[j] + d) % N;
      if (a > b) swap(a, b);
      if (se.find(make_pair(a, b)) == se.end()) {
        ng = true;
        break;
      }
    }
    if (ng == false) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
  return 0;
}

```