#include <bits/stdc++.h>
using namespace std;
template <typename T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;
const long long MOD = 1e9 + 7;
int N, M;
set<int> ep[100100];
set<int> l;
vector<int> dd;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  int n = N;
  for (int d = 2; d * d <= N; d++) {
    if (n % d) continue;
    dd.push_back(N / d);
    while (n % d == 0) n /= d;
  }
  if (n > 1) dd.push_back(N / n);
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    int k = (b - a + N) % N;
    if (2 * k < N) {
      ep[k].insert(a - 1);
      l.insert(k);
    } else if (2 * k > N) {
      ep[N - k].insert(b - 1);
      l.insert(N - k);
    } else {
      ep[k].insert(a - 1);
      ep[k].insert(b - 1);
      l.insert(k);
    }
  }
  for (int k : dd) {
    bool gd = 1;
    for (int r : l) {
      auto& s = ep[r];
      bool good = 1;
      for (int a : s) {
        if (!s.count((a + k) % N)) {
          good = 0;
          break;
        }
      }
      if (!good) {
        gd = 0;
        break;
      }
    }
    if (gd) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No\n";
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
template <typename T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;
const long long MOD = 1e9 + 7;
int N, M;
set<int> ep[100100];
set<int> l;
vector<int> dd;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  int n = N;
  for (int d = 2; d * d <= N; d++) {
    if (n % d) continue;
    dd.push_back(N / d);
    while (n % d == 0) n /= d;
  }
  if (n > 1) dd.push_back(N / n);
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    int k = (b - a + N) % N;
    if (2 * k < N) {
      ep[k].insert(a - 1);
      l.insert(k);
    } else if (2 * k > N) {
      ep[N - k].insert(b - 1);
      l.insert(N - k);
    } else {
      ep[k].insert(a - 1);
      ep[k].insert(b - 1);
      l.insert(k);
    }
  }
  for (int k : dd) {
    bool gd = 1;
    for (int r : l) {
      auto& s = ep[r];
      bool good = 1;
      for (int a : s) {
        if (!s.count((a + k) % N)) {
          good = 0;
          break;
        }
      }
      if (!good) {
        gd = 0;
        break;
      }
    }
    if (gd) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No\n";
  return 0;
}

```