#include <bits/stdc++.h>
using namespace std;
int GCD(int a, int b) {
  if (b == 0) return a;
  return GCD(b, a % b);
}
int main() {
  int n, m, i, j, u, v;
  cin >> n >> m;
  vector<vector<int> > cnt(n + 1);
  vector<set<int> > At_this(n + 1);
  for (i = 0; i < m; i++) {
    cin >> u >> v;
    if (v > u) {
      j = v;
      v = u;
      u = j;
    }
    j = u - v;
    cnt[j].push_back(v);
    cnt[n - j].push_back(u);
    At_this[v].insert(j);
    At_this[u].insert(n - j);
  }
  int GG = n;
  for (i = 1; i <= n; i++) {
    if (cnt[i].size() > 0) {
      j = GCD(n, (int)cnt[i].size());
      GG = GCD(GG, j);
    }
  }
  GG = n / GG;
  int flag = 1;
  for (i = 1; i <= n; i++) {
    for (j = 0; j < cnt[i].size(); j++) {
      u = cnt[i][j];
      u = (u + GG);
      if (u > n) u -= n;
      if (At_this[u].find(i) == At_this[u].end()) {
        flag = 0;
      }
    }
  }
  if (GG == n)
    cout << "No" << endl;
  else {
    if (flag)
      cout << "Yes" << endl;
    else
      cout << "No" << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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 GCD(int a, int b) {
  if (b == 0) return a;
  return GCD(b, a % b);
}
int main() {
  int n, m, i, j, u, v;
  cin >> n >> m;
  vector<vector<int> > cnt(n + 1);
  vector<set<int> > At_this(n + 1);
  for (i = 0; i < m; i++) {
    cin >> u >> v;
    if (v > u) {
      j = v;
      v = u;
      u = j;
    }
    j = u - v;
    cnt[j].push_back(v);
    cnt[n - j].push_back(u);
    At_this[v].insert(j);
    At_this[u].insert(n - j);
  }
  int GG = n;
  for (i = 1; i <= n; i++) {
    if (cnt[i].size() > 0) {
      j = GCD(n, (int)cnt[i].size());
      GG = GCD(GG, j);
    }
  }
  GG = n / GG;
  int flag = 1;
  for (i = 1; i <= n; i++) {
    for (j = 0; j < cnt[i].size(); j++) {
      u = cnt[i][j];
      u = (u + GG);
      if (u > n) u -= n;
      if (At_this[u].find(i) == At_this[u].end()) {
        flag = 0;
      }
    }
  }
  if (GG == n)
    cout << "No" << endl;
  else {
    if (flag)
      cout << "Yes" << endl;
    else
      cout << "No" << endl;
  }
  return 0;
}

```