#include <bits/stdc++.h>
using namespace std;
int n, vi[100002], fl;
set<int, greater<int> > pq[100002];
int fn2(int p) {
  if (p > n) return p - n;
  return p;
}
void fn(int p) {
  int i, fl2 = 1;
  for (i = p * p; i <= (n); i += p * 2) vi[i]++;
  p = n / p;
  for (i = 1; i <= n; i++) {
    if (pq[i] != pq[fn2(i + p)]) {
      fl2 = 0;
      break;
    }
  }
  if (fl2) fl = 1;
}
int main() {
  int m, a, b, i;
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    if (a > b) swap(a, b);
    pq[a].insert(b - a);
    pq[b].insert(n - b + a);
  }
  if (n % 2 == 0) {
    fn(2);
  }
  if (fl) {
    cout << "Yes";
    return 0;
  }
  for (i = 3; i <= (n); i += 2) {
    if (vi[i]) continue;
    if (n % i == 0) {
      fn(i);
    }
    if (fl) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
  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 n, vi[100002], fl;
set<int, greater<int> > pq[100002];
int fn2(int p) {
  if (p > n) return p - n;
  return p;
}
void fn(int p) {
  int i, fl2 = 1;
  for (i = p * p; i <= (n); i += p * 2) vi[i]++;
  p = n / p;
  for (i = 1; i <= n; i++) {
    if (pq[i] != pq[fn2(i + p)]) {
      fl2 = 0;
      break;
    }
  }
  if (fl2) fl = 1;
}
int main() {
  int m, a, b, i;
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    if (a > b) swap(a, b);
    pq[a].insert(b - a);
    pq[b].insert(n - b + a);
  }
  if (n % 2 == 0) {
    fn(2);
  }
  if (fl) {
    cout << "Yes";
    return 0;
  }
  for (i = 3; i <= (n); i += 2) {
    if (vi[i]) continue;
    if (n % i == 0) {
      fn(i);
    }
    if (fl) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
  return 0;
}

```