#include <bits/stdc++.h>
using namespace std;
long long maxi = LLONG_MIN;
long long mini = LLONG_MAX;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long t, i, j, k, n, temp, flag, ans, m;
vector<long long> a[200005];
int main() {
  int n, k;
  cin >> n >> k;
  vector<long long> f;
  for (int i = 1; i <= k; i++) {
    int x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) sort(a[i].begin(), a[i].end());
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      f.push_back(i);
      if (i * i != n) {
        f.push_back(n / i);
      }
    }
  }
  f.push_back(1);
  sort(f.begin(), f.end());
  for (int i = 0; i < f.size(); i++) {
    int d = 0;
    for (int j = 1; j <= n; j++) {
      vector<long long> x;
      for (int l = 0; l < a[j].size(); l++) {
        x.push_back((a[j][l] + f[i] - 1) % n + 1);
      }
      int h = (j + f[i] - 1) % n + 1;
      sort(x.begin(), x.end());
      if (x.size() != a[h].size()) {
        break;
      }
      int c = 0;
      for (int l = 0; l < a[h].size() && flag == 0; l++) {
        if (a[h][l] == x[l]) c++;
      }
      if (c != a[h].size()) {
        break;
      } else {
        d++;
      }
    }
    if (d == n) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
long long maxi = LLONG_MIN;
long long mini = LLONG_MAX;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long t, i, j, k, n, temp, flag, ans, m;
vector<long long> a[200005];
int main() {
  int n, k;
  cin >> n >> k;
  vector<long long> f;
  for (int i = 1; i <= k; i++) {
    int x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) sort(a[i].begin(), a[i].end());
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      f.push_back(i);
      if (i * i != n) {
        f.push_back(n / i);
      }
    }
  }
  f.push_back(1);
  sort(f.begin(), f.end());
  for (int i = 0; i < f.size(); i++) {
    int d = 0;
    for (int j = 1; j <= n; j++) {
      vector<long long> x;
      for (int l = 0; l < a[j].size(); l++) {
        x.push_back((a[j][l] + f[i] - 1) % n + 1);
      }
      int h = (j + f[i] - 1) % n + 1;
      sort(x.begin(), x.end());
      if (x.size() != a[h].size()) {
        break;
      }
      int c = 0;
      for (int l = 0; l < a[h].size() && flag == 0; l++) {
        if (a[h][l] == x[l]) c++;
      }
      if (c != a[h].size()) {
        break;
      } else {
        d++;
      }
    }
    if (d == n) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
  return 0;
}

```