#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> check;
  check.push_back(1);
  for (long long i = 2; i < n; i++) {
    if (isPrime(i) && n % i == 0) {
      check.push_back(n / i);
    }
  }
  pair<long long, long long> p[m];
  for (long long i = 0; i < m; i++) {
    long long u, v;
    cin >> u >> v;
    u--;
    v--;
    if (u > v) {
      swap(u, v);
    }
    p[i] = {u, v};
  }
  sort(p, p + m);
  for (auto x : check) {
    pair<long long, long long> p2[m];
    for (long long i = 0; i < m; i++) {
      long long a = (p[i].first + x) % n;
      long long b = (p[i].second + x) % n;
      if (a > b) {
        swap(a, b);
      }
      p2[i] = {a, b};
    }
    sort(p2, p2 + m);
    long long flag = 1;
    for (long long i = 0; i < m; i++) {
      if (p2[i] != p[i]) {
        flag = 0;
        break;
      }
    }
    if (flag) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No\n";
  return 0;
}

### Prompt

Generate 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;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> check;
  check.push_back(1);
  for (long long i = 2; i < n; i++) {
    if (isPrime(i) && n % i == 0) {
      check.push_back(n / i);
    }
  }
  pair<long long, long long> p[m];
  for (long long i = 0; i < m; i++) {
    long long u, v;
    cin >> u >> v;
    u--;
    v--;
    if (u > v) {
      swap(u, v);
    }
    p[i] = {u, v};
  }
  sort(p, p + m);
  for (auto x : check) {
    pair<long long, long long> p2[m];
    for (long long i = 0; i < m; i++) {
      long long a = (p[i].first + x) % n;
      long long b = (p[i].second + x) % n;
      if (a > b) {
        swap(a, b);
      }
      p2[i] = {a, b};
    }
    sort(p2, p2 + m);
    long long flag = 1;
    for (long long i = 0; i < m; i++) {
      if (p2[i] != p[i]) {
        flag = 0;
        break;
      }
    }
    if (flag) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No\n";
  return 0;
}

```