#include <bits/stdc++.h>
using namespace std;
bool check(int i, int n, set<pair<int, int> >& st) {
  for (auto it : st) {
    pair<int, int> t(it.first + i, it.second + i);
    if (t.first > n) t.first %= n;
    if (t.second > n) t.second %= n;
    if (t.first > t.second) swap(t.first, t.second);
    if (!st.count(t)) return false;
  }
  return true;
}
int main() {
  int n, m, a, b;
  while (cin >> n >> m) {
    set<pair<int, int> > st;
    for (int i = 1; i <= m; i++) {
      cin >> a >> b;
      st.insert(make_pair(min(a, b), max(a, b)));
    }
    bool flg = false;
    for (int i = 1; i * i <= n; i++) {
      if (n % i == 0) {
        if (check(i, n, st)) {
          flg = true;
          break;
        } else if (i != 1 && check(n / i, n, st)) {
          flg = true;
          break;
        }
      }
    }
    cout << (flg ? "Yes" : "No") << endl;
  }
  return 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;
bool check(int i, int n, set<pair<int, int> >& st) {
  for (auto it : st) {
    pair<int, int> t(it.first + i, it.second + i);
    if (t.first > n) t.first %= n;
    if (t.second > n) t.second %= n;
    if (t.first > t.second) swap(t.first, t.second);
    if (!st.count(t)) return false;
  }
  return true;
}
int main() {
  int n, m, a, b;
  while (cin >> n >> m) {
    set<pair<int, int> > st;
    for (int i = 1; i <= m; i++) {
      cin >> a >> b;
      st.insert(make_pair(min(a, b), max(a, b)));
    }
    bool flg = false;
    for (int i = 1; i * i <= n; i++) {
      if (n % i == 0) {
        if (check(i, n, st)) {
          flg = true;
          break;
        } else if (i != 1 && check(n / i, n, st)) {
          flg = true;
          break;
        }
      }
    }
    cout << (flg ? "Yes" : "No") << endl;
  }
  return 0;
}

```