#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, mod = 1e9 + 7, inf = 1e9 + 7;
int n, m;
pair<int, int> a[N];
map<pair<int, int>, int> mpp;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a[i].first >> a[i].second;
    mpp[{--a[i].first, --a[i].second}] = 1;
  }
  int s = sqrt(n);
  for (int i = 1; i <= s; i++) {
    if (n % i == 0) {
      bool ck = 1;
      for (int j = 0; j < m; j++) {
        if (!mpp[{(a[j].first + i) % n, (a[j].second + i) % n}] &&
            !mpp[{(a[j].second + i) % n, (a[j].first + i) % n}]) {
          ck = 0;
          break;
        }
      }
      if (ck) return cout << "Yes", 0;
      ;
      ck = 1;
      for (int j = 0; j < m; j++) {
        if (!mpp[{(a[j].first + n / i) % n, (a[j].second + n / i) % n}] &&
            !mpp[{(a[j].second + n / i) % n, (a[j].first + n / i) % n}]) {
          ck = 0;
          break;
        }
      }
      if (ck && i != 1) return cout << "Yes", 0;
      ;
    }
  }
  cout << "No";
}

### Prompt

Create a solution in Cpp for 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;
const int N = 2e5 + 5, mod = 1e9 + 7, inf = 1e9 + 7;
int n, m;
pair<int, int> a[N];
map<pair<int, int>, int> mpp;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a[i].first >> a[i].second;
    mpp[{--a[i].first, --a[i].second}] = 1;
  }
  int s = sqrt(n);
  for (int i = 1; i <= s; i++) {
    if (n % i == 0) {
      bool ck = 1;
      for (int j = 0; j < m; j++) {
        if (!mpp[{(a[j].first + i) % n, (a[j].second + i) % n}] &&
            !mpp[{(a[j].second + i) % n, (a[j].first + i) % n}]) {
          ck = 0;
          break;
        }
      }
      if (ck) return cout << "Yes", 0;
      ;
      ck = 1;
      for (int j = 0; j < m; j++) {
        if (!mpp[{(a[j].first + n / i) % n, (a[j].second + n / i) % n}] &&
            !mpp[{(a[j].second + n / i) % n, (a[j].first + n / i) % n}]) {
          ck = 0;
          break;
        }
      }
      if (ck && i != 1) return cout << "Yes", 0;
      ;
    }
  }
  cout << "No";
}

```