#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  pair<int, int> b[m];
  map<pair<int, int>, int> d;
  int half = 0;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    half += abs(u - v) == n / 2;
    if (u > v) swap(u, v);
    if (n + u - v < v - u) swap(u, v);
    b[i].first = u, b[i].second = v;
    b[i].first--, b[i].second--;
    d[b[i]]++;
  }
  if (n % 2 == 0 && half == m) {
    cout << "Yes";
    return 0;
  }
  for (int i = 2; i * i <= n; i++) {
    if (n % i) continue;
    int j = i;
    bool cs = true;
    for (auto e : b) {
      if (d[make_pair((e.first + j) % n, (e.second + j) % n)] == 0) {
        cs = false;
        break;
      }
    }
    if (cs) {
      cout << "Yes";
      return 0;
    }
    j = n / i;
    cs = true;
    for (auto e : b) {
      if (d[make_pair((e.first + j) % n, (e.second + j) % n)] == 0) {
        cs = false;
        break;
      }
    }
    if (cs) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
}

### Prompt

Develop a solution in CPP 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  pair<int, int> b[m];
  map<pair<int, int>, int> d;
  int half = 0;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    half += abs(u - v) == n / 2;
    if (u > v) swap(u, v);
    if (n + u - v < v - u) swap(u, v);
    b[i].first = u, b[i].second = v;
    b[i].first--, b[i].second--;
    d[b[i]]++;
  }
  if (n % 2 == 0 && half == m) {
    cout << "Yes";
    return 0;
  }
  for (int i = 2; i * i <= n; i++) {
    if (n % i) continue;
    int j = i;
    bool cs = true;
    for (auto e : b) {
      if (d[make_pair((e.first + j) % n, (e.second + j) % n)] == 0) {
        cs = false;
        break;
      }
    }
    if (cs) {
      cout << "Yes";
      return 0;
    }
    j = n / i;
    cs = true;
    for (auto e : b) {
      if (d[make_pair((e.first + j) % n, (e.second + j) % n)] == 0) {
        cs = false;
        break;
      }
    }
    if (cs) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
}

```