#include <bits/stdc++.h>
using namespace std;
vector<long long> find(long long n) {
  vector<long long> ans;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ans.push_back(i);
      if (i == 1) continue;
      if (n / i != i) ans.push_back(n / i);
    }
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  pair<long long, long long> arr[m];
  for (long long i = 0; i < m; i++) {
    cin >> arr[i].first >> arr[i].second;
    if (arr[i].first > arr[i].second) swap(arr[i].first, arr[i].second);
  }
  map<pair<long long, long long>, long long> mp;
  for (long long i = 0; i < m; i++) {
    mp[arr[i]] = 1;
  }
  vector<long long> a = find(n);
  for (long long i = 0; i < a.size(); i++) {
    long long cnt = 0;
    for (long long j = 0; j < m; j++) {
      long long x = arr[j].first + a[i];
      x %= n;
      if (x == 0) x += n;
      long long y = arr[j].second + a[i];
      y %= n;
      if (y == 0) y += n;
      if (x > y) swap(x, y);
      if (mp[{x, y}] == 0) break;
      cnt++;
    }
    if (cnt == m) {
      cout << "YES" << endl;
      return 0;
    }
  }
  cout << "NO" << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
vector<long long> find(long long n) {
  vector<long long> ans;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ans.push_back(i);
      if (i == 1) continue;
      if (n / i != i) ans.push_back(n / i);
    }
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  pair<long long, long long> arr[m];
  for (long long i = 0; i < m; i++) {
    cin >> arr[i].first >> arr[i].second;
    if (arr[i].first > arr[i].second) swap(arr[i].first, arr[i].second);
  }
  map<pair<long long, long long>, long long> mp;
  for (long long i = 0; i < m; i++) {
    mp[arr[i]] = 1;
  }
  vector<long long> a = find(n);
  for (long long i = 0; i < a.size(); i++) {
    long long cnt = 0;
    for (long long j = 0; j < m; j++) {
      long long x = arr[j].first + a[i];
      x %= n;
      if (x == 0) x += n;
      long long y = arr[j].second + a[i];
      y %= n;
      if (y == 0) y += n;
      if (x > y) swap(x, y);
      if (mp[{x, y}] == 0) break;
      cnt++;
    }
    if (cnt == m) {
      cout << "YES" << endl;
      return 0;
    }
  }
  cout << "NO" << endl;
  return 0;
}

```