#include <bits/stdc++.h>
using namespace std;
int n;
inline vector<vector<int>> f(const vector<pair<int, int>> &edges,
                             const int offset) {
  cerr << "offset"
       << " = " << (offset) << endl;
  vector<vector<int>> adj(n);
  for (const pair<int, int> &e : edges) {
    int u = e.first;
    int v = e.second;
    u += offset;
    v += offset;
    if (u >= n) u -= n;
    if (v >= n) v -= n;
    assert(0 <= u && u < n);
    assert(0 <= v && v < n);
    if (u > v) swap(u, v);
    adj[u].push_back(v);
  }
  for (int i = 0; i < n; ++i) sort(adj[i].begin(), adj[i].end());
  return adj;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int m;
  cin >> n >> m;
  vector<pair<int, int>> edges(m);
  for (auto &&e : edges) cin >> e.first >> e.second;
  const vector<vector<int>> adj = f(edges, 0);
  for (int i = 1; i < n; ++i)
    if (n % i == 0 && f(edges, i) == adj) {
      cout << "Yes";
      return 0;
    }
  cout << "No";
  return 0;
}

### Prompt

Create a solution in CPP for 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 n;
inline vector<vector<int>> f(const vector<pair<int, int>> &edges,
                             const int offset) {
  cerr << "offset"
       << " = " << (offset) << endl;
  vector<vector<int>> adj(n);
  for (const pair<int, int> &e : edges) {
    int u = e.first;
    int v = e.second;
    u += offset;
    v += offset;
    if (u >= n) u -= n;
    if (v >= n) v -= n;
    assert(0 <= u && u < n);
    assert(0 <= v && v < n);
    if (u > v) swap(u, v);
    adj[u].push_back(v);
  }
  for (int i = 0; i < n; ++i) sort(adj[i].begin(), adj[i].end());
  return adj;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int m;
  cin >> n >> m;
  vector<pair<int, int>> edges(m);
  for (auto &&e : edges) cin >> e.first >> e.second;
  const vector<vector<int>> adj = f(edges, 0);
  for (int i = 1; i < n; ++i)
    if (n % i == 0 && f(edges, i) == adj) {
      cout << "Yes";
      return 0;
    }
  cout << "No";
  return 0;
}

```