#include <bits/stdc++.h>
using namespace std;
int n, m, sz, d[1001], a[200002], b[200002];
set<pair<int, int> > s;
void div() {
  sz = 0;
  for (int i = 1; i <= trunc(sqrt(n)); i++)
    if (n % i == 0) {
      d[++sz] = i;
      if (i * i != n && i != 1) d[++sz] = n / i;
    }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i] >> b[i];
    s.insert(pair<int, int>(a[i], b[i]));
    s.insert(pair<int, int>(b[i], a[i]));
  }
  div();
  bool check;
  for (int i = 1; i <= sz; i++) {
    check = 1;
    for (int j = 1; j <= m; j++) {
      int x = a[j] + d[i];
      int y = b[j] + d[i];
      if (x > n) x -= n;
      if (y > n) y -= n;
      if (!s.count(pair<int, int>(x, y))) check = 0;
    }
    if (check) return cout << "YES", 0;
  }
  cout << "NO";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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, m, sz, d[1001], a[200002], b[200002];
set<pair<int, int> > s;
void div() {
  sz = 0;
  for (int i = 1; i <= trunc(sqrt(n)); i++)
    if (n % i == 0) {
      d[++sz] = i;
      if (i * i != n && i != 1) d[++sz] = n / i;
    }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i] >> b[i];
    s.insert(pair<int, int>(a[i], b[i]));
    s.insert(pair<int, int>(b[i], a[i]));
  }
  div();
  bool check;
  for (int i = 1; i <= sz; i++) {
    check = 1;
    for (int j = 1; j <= m; j++) {
      int x = a[j] + d[i];
      int y = b[j] + d[i];
      if (x > n) x -= n;
      if (y > n) y -= n;
      if (!s.count(pair<int, int>(x, y))) check = 0;
    }
    if (check) return cout << "YES", 0;
  }
  cout << "NO";
  return 0;
}

```