#include <bits/stdc++.h>
using namespace std;
const int M = 2e5 + 23;
int a[M], b[M], n, m;
map<int, bool> mapa[M];
void proveri(int k) {
  bool ok = 1;
  for (int i = 1; i <= m; i++) {
    int a1 = (a[i] + k - 1) % n + 1, b1 = (b[i] + k - 1) % n + 1;
    if (!mapa[a1].count(b1)) {
      ok = 0;
      break;
    }
  }
  if (ok) {
    cout << "Yes" << endl;
    exit(0);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i] >> b[i];
    mapa[a[i]][b[i]] = 1;
    mapa[b[i]][a[i]] = 1;
  }
  for (int i = 1; i * i <= n; i++) {
    if (n % i) {
      continue;
    }
    proveri(i);
    if (i != 1) {
      proveri(n / i);
    }
  }
  cout << "No" << endl;
}

### 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;
const int M = 2e5 + 23;
int a[M], b[M], n, m;
map<int, bool> mapa[M];
void proveri(int k) {
  bool ok = 1;
  for (int i = 1; i <= m; i++) {
    int a1 = (a[i] + k - 1) % n + 1, b1 = (b[i] + k - 1) % n + 1;
    if (!mapa[a1].count(b1)) {
      ok = 0;
      break;
    }
  }
  if (ok) {
    cout << "Yes" << endl;
    exit(0);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i] >> b[i];
    mapa[a[i]][b[i]] = 1;
    mapa[b[i]][a[i]] = 1;
  }
  for (int i = 1; i * i <= n; i++) {
    if (n % i) {
      continue;
    }
    proveri(i);
    if (i != 1) {
      proveri(n / i);
    }
  }
  cout << "No" << endl;
}

```