#include <bits/stdc++.h>
using namespace std;
vector<int> adj[200005];
vector<int> lps(vector<int> v) {
  vector<int> pref_arr(v.size());
  pref_arr[0] = 0;
  int i = 1, j = 0;
  for (; i < v.size();) {
    if (v[i] == v[j])
      pref_arr[i++] = ++j;
    else {
      if (j == 0)
        pref_arr[i++] = 0;
      else
        j = pref_arr[j - 1];
    }
  }
  return pref_arr;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    int n, k, i, j;
    cin >> n >> k;
    vector<pair<int, int> > v;
    for (i = 0; i < k; i++) {
      int x, y;
      cin >> x >> y;
      adj[x].push_back((n + y - x) % n);
      adj[y].push_back((n + x - y) % n);
    }
    for (i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
    vector<int> s;
    for (i = 1; i <= n; i++) {
      if (adj[i].size()) {
        for (auto j : adj[i]) s.push_back(j);
      } else {
        s.push_back(-1);
      }
    }
    vector<int> temp = s;
    for (auto j : s) temp.push_back(j);
    vector<int> pref_arr = lps(temp);
    for (i = s.size(); i < temp.size() - 1; i++) {
      if (pref_arr[i] == s.size()) {
        cout << "Yes";
        return 0;
      }
    }
    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;
vector<int> adj[200005];
vector<int> lps(vector<int> v) {
  vector<int> pref_arr(v.size());
  pref_arr[0] = 0;
  int i = 1, j = 0;
  for (; i < v.size();) {
    if (v[i] == v[j])
      pref_arr[i++] = ++j;
    else {
      if (j == 0)
        pref_arr[i++] = 0;
      else
        j = pref_arr[j - 1];
    }
  }
  return pref_arr;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    int n, k, i, j;
    cin >> n >> k;
    vector<pair<int, int> > v;
    for (i = 0; i < k; i++) {
      int x, y;
      cin >> x >> y;
      adj[x].push_back((n + y - x) % n);
      adj[y].push_back((n + x - y) % n);
    }
    for (i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
    vector<int> s;
    for (i = 1; i <= n; i++) {
      if (adj[i].size()) {
        for (auto j : adj[i]) s.push_back(j);
      } else {
        s.push_back(-1);
      }
    }
    vector<int> temp = s;
    for (auto j : s) temp.push_back(j);
    vector<int> pref_arr = lps(temp);
    for (i = s.size(); i < temp.size() - 1; i++) {
      if (pref_arr[i] == s.size()) {
        cout << "Yes";
        return 0;
      }
    }
    cout << "No";
  }
  return 0;
}

```