#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > l;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a %= n;
    b %= n;
    l.push_back(make_pair(max(a, b), min(a, b)));
  }
  sort(l.begin(), l.end());
  int t = 0;
  set<int> fact;
  for (int i = 2; i < (int)sqrt(n) + 1; i++) {
    if (n % i == 0) {
      fact.insert(i);
      fact.insert(n / i);
    }
  }
  fact.insert(1);
  for (auto &fi : fact) {
    vector<pair<int, int> > l1;
    int f = 0;
    for (int i = 0; i < m; i++) {
      int a = (l[i].first + fi) % n;
      int b = (l[i].second + fi) % n;
      l1.push_back(make_pair(max(a, b), min(a, b)));
    }
    sort(l1.begin(), l1.end());
    for (int i = 0; i < m; i++) {
      if ((l1[i].first != l[i].first) || (l1[i].second != l[i].second)) {
        f = 1;
        break;
      }
    }
    if (f == 0) {
      t = 1;
      cout << "Yes" << endl;
      break;
    }
  }
  if (t == 0) cout << "No" << endl;
}
int main() {
  cin.sync_with_stdio(false), cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  int c = 0;
  while (t--) {
    solve();
    c += 1;
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
const int N = 100005;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > l;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a %= n;
    b %= n;
    l.push_back(make_pair(max(a, b), min(a, b)));
  }
  sort(l.begin(), l.end());
  int t = 0;
  set<int> fact;
  for (int i = 2; i < (int)sqrt(n) + 1; i++) {
    if (n % i == 0) {
      fact.insert(i);
      fact.insert(n / i);
    }
  }
  fact.insert(1);
  for (auto &fi : fact) {
    vector<pair<int, int> > l1;
    int f = 0;
    for (int i = 0; i < m; i++) {
      int a = (l[i].first + fi) % n;
      int b = (l[i].second + fi) % n;
      l1.push_back(make_pair(max(a, b), min(a, b)));
    }
    sort(l1.begin(), l1.end());
    for (int i = 0; i < m; i++) {
      if ((l1[i].first != l[i].first) || (l1[i].second != l[i].second)) {
        f = 1;
        break;
      }
    }
    if (f == 0) {
      t = 1;
      cout << "Yes" << endl;
      break;
    }
  }
  if (t == 0) cout << "No" << endl;
}
int main() {
  cin.sync_with_stdio(false), cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  int c = 0;
  while (t--) {
    solve();
    c += 1;
  }
  return 0;
}

```