#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  set<int> factlist;
  long double cr = n;
  long double uf = ceil(sqrt(cr)) + 1;
  for (int i = 1; i < uf; i++) {
    if (n % i == 0) {
      factlist.insert(i);
      factlist.insert(n / i);
    }
  }
  vector<int> flist;
  for (set<int>::iterator l = factlist.begin(); l != factlist.end(); ++l) {
    if (*l != 0) {
      flist.push_back(*l);
    }
  }
  vector<pair<int, int> > original;
  for (int i = 0; i < k; i++) {
    int a, b;
    cin >> a >> b;
    int a_, b_;
    a_ = min(a, b);
    b_ = max(a, b);
    a = a_;
    b = b_;
    original.push_back(make_pair(a - 1, b - 1));
  }
  sort(original.begin(), original.end());
  for (int i = 0; i < flist.size() - 1; i++) {
    int currf = flist[i];
    vector<pair<int, int> > nst;
    for (int i = 0; i < k; i++) {
      int a = (original[i].first + currf) % n;
      int b = (original[i].second + currf) % n;
      int _a = min(a, b), _b = max(a, b);
      a = _a;
      b = _b;
      nst.push_back(make_pair(_a, _b));
    }
    sort(nst.begin(), nst.end());
    bool can = true;
    for (int i = 0; i < nst.size() && can; i++) {
      if (nst[i].first == original[i].first &&
          nst[i].second == original[i].second) {
      } else {
        can = false;
      }
    }
    if (can) {
      cout << "Yes";
      exit(0);
    }
  }
  cout << "No";
  exit(0);
}

### 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;
int main() {
  int n, k;
  cin >> n >> k;
  set<int> factlist;
  long double cr = n;
  long double uf = ceil(sqrt(cr)) + 1;
  for (int i = 1; i < uf; i++) {
    if (n % i == 0) {
      factlist.insert(i);
      factlist.insert(n / i);
    }
  }
  vector<int> flist;
  for (set<int>::iterator l = factlist.begin(); l != factlist.end(); ++l) {
    if (*l != 0) {
      flist.push_back(*l);
    }
  }
  vector<pair<int, int> > original;
  for (int i = 0; i < k; i++) {
    int a, b;
    cin >> a >> b;
    int a_, b_;
    a_ = min(a, b);
    b_ = max(a, b);
    a = a_;
    b = b_;
    original.push_back(make_pair(a - 1, b - 1));
  }
  sort(original.begin(), original.end());
  for (int i = 0; i < flist.size() - 1; i++) {
    int currf = flist[i];
    vector<pair<int, int> > nst;
    for (int i = 0; i < k; i++) {
      int a = (original[i].first + currf) % n;
      int b = (original[i].second + currf) % n;
      int _a = min(a, b), _b = max(a, b);
      a = _a;
      b = _b;
      nst.push_back(make_pair(_a, _b));
    }
    sort(nst.begin(), nst.end());
    bool can = true;
    for (int i = 0; i < nst.size() && can; i++) {
      if (nst[i].first == original[i].first &&
          nst[i].second == original[i].second) {
      } else {
        can = false;
      }
    }
    if (can) {
      cout << "Yes";
      exit(0);
    }
  }
  cout << "No";
  exit(0);
}

```