#include <bits/stdc++.h>
using namespace std;
long long n, t, r, maxn, x, maxi = 0LL, flag = 0, l1, r1, l2, r2,
                            parity[2] = {0, 0}, ok, q;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  t = 1;
  long long uu = t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> v;
    map<pair<long long, long long>, long long> mpp;
    vector<pair<long long, long long>> b;
    for (long long i = 0; i < m; i++) {
      long long x, y;
      cin >> x >> y;
      x--, y--;
      long long c = x + y;
      x = min(x, y), y = c - x;
      b.push_back(make_pair(x, y));
      mpp[make_pair(x, y)]++;
    }
    vector<long long> a;
    for (long long i = 1; i <= sqrt(n) + 1; i++)
      if (n % i == 0) {
        if (i != n / i) a.push_back(i);
        a.push_back(n / i);
      }
    sort(a.begin(), a.end());
    for (long long i = 0; i < a.size(); i++) {
      if (a[i] >= n) break;
      long long flag = 0;
      for (long long j = 0; j < m; j++) {
        long long x = (b[j].first + a[i]) % n, y = (b[j].second + a[i]) % n;
        long long c = x + y;
        x = min(x, y), y = c - x;
        if (mpp[make_pair(x, y)] == 0) {
          flag = 1;
          break;
        }
      }
      if (flag == 0) {
        cout << "Yes";
        return 0;
      }
    }
    cout << ((n == 1) ? "Yes" : "No");
  }
  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;
long long n, t, r, maxn, x, maxi = 0LL, flag = 0, l1, r1, l2, r2,
                            parity[2] = {0, 0}, ok, q;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  t = 1;
  long long uu = t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> v;
    map<pair<long long, long long>, long long> mpp;
    vector<pair<long long, long long>> b;
    for (long long i = 0; i < m; i++) {
      long long x, y;
      cin >> x >> y;
      x--, y--;
      long long c = x + y;
      x = min(x, y), y = c - x;
      b.push_back(make_pair(x, y));
      mpp[make_pair(x, y)]++;
    }
    vector<long long> a;
    for (long long i = 1; i <= sqrt(n) + 1; i++)
      if (n % i == 0) {
        if (i != n / i) a.push_back(i);
        a.push_back(n / i);
      }
    sort(a.begin(), a.end());
    for (long long i = 0; i < a.size(); i++) {
      if (a[i] >= n) break;
      long long flag = 0;
      for (long long j = 0; j < m; j++) {
        long long x = (b[j].first + a[i]) % n, y = (b[j].second + a[i]) % n;
        long long c = x + y;
        x = min(x, y), y = c - x;
        if (mpp[make_pair(x, y)] == 0) {
          flag = 1;
          break;
        }
      }
      if (flag == 0) {
        cout << "Yes";
        return 0;
      }
    }
    cout << ((n == 1) ? "Yes" : "No");
  }
  return 0;
}

```