#include <bits/stdc++.h>
using namespace std;
using ll = long long;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> start(m);
  for (auto &el : start) {
    cin >> el.first >> el.second;
    el.first--;
    el.second--;
    if (el.first > el.second) {
      swap(el.first, el.second);
    }
  }
  sort(start.begin(), start.end());
  function<bool(int)> Check = [&](int del) {
    vector<pair<int, int>> cur = start;
    for (auto &el : cur) {
      el.first += del;
      el.second += del;
      if (el.first >= n) {
        el.first -= n;
      }
      if (el.second >= n) {
        el.second -= n;
      }
      if (el.first > el.second) {
        swap(el.first, el.second);
      }
    }
    sort(cur.begin(), cur.end());
    return (cur == start);
  };
  bool fnd = false;
  vector<int> is_prime(n + 1, 1);
  for (int i = 2; i <= n; i++) {
    if (is_prime[i]) {
      for (int j = i + i; j <= n; j += i) {
        is_prime[j] = 0;
      }
    }
  }
  for (int i = 2; i <= n; i++) {
    if (is_prime[i] && n % i == 0) {
      fnd |= Check(n / i);
    }
  }
  cout << (fnd ? "YES\n" : "NO\n");
}

### Prompt

Please provide a CPP coded solution 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;
using ll = long long;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> start(m);
  for (auto &el : start) {
    cin >> el.first >> el.second;
    el.first--;
    el.second--;
    if (el.first > el.second) {
      swap(el.first, el.second);
    }
  }
  sort(start.begin(), start.end());
  function<bool(int)> Check = [&](int del) {
    vector<pair<int, int>> cur = start;
    for (auto &el : cur) {
      el.first += del;
      el.second += del;
      if (el.first >= n) {
        el.first -= n;
      }
      if (el.second >= n) {
        el.second -= n;
      }
      if (el.first > el.second) {
        swap(el.first, el.second);
      }
    }
    sort(cur.begin(), cur.end());
    return (cur == start);
  };
  bool fnd = false;
  vector<int> is_prime(n + 1, 1);
  for (int i = 2; i <= n; i++) {
    if (is_prime[i]) {
      for (int j = i + i; j <= n; j += i) {
        is_prime[j] = 0;
      }
    }
  }
  for (int i = 2; i <= n; i++) {
    if (is_prime[i] && n % i == 0) {
      fnd |= Check(n / i);
    }
  }
  cout << (fnd ? "YES\n" : "NO\n");
}

```