#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
void solve();
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(10);
  int t = 1;
  while (t--) solve();
  return 0;
}
int modd(int a, int b) {
  a = a % b;
  if (a < 0) a = a + b;
  return a;
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > s;
  for (int i = 0; i < (int)m; i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    if (x > y) swap(x, y);
    s.push_back({x, y});
  }
  sort((s).begin(), (s).end());
  for (int i = 2; i < (int)n + 1; i++) {
    if (n % i != 0) continue;
    int t = n / i;
    for (int d : {t}) {
      vector<pair<int, int> > f;
      for (auto p : s) {
        int e1 = modd(p.first + d, n);
        int e2 = modd(p.second + d, n);
        if (e1 > e2) swap(e1, e2);
        f.push_back({e1, e2});
      }
      sort((f).begin(), (f).end());
      if (s == f) {
        cout << "Yes"
             << "\n";
        return;
      }
    }
  }
  {
    cout << "No"
         << "\n";
    return;
  };
}

### Prompt

Please create a solution in CPP 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;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
void solve();
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(10);
  int t = 1;
  while (t--) solve();
  return 0;
}
int modd(int a, int b) {
  a = a % b;
  if (a < 0) a = a + b;
  return a;
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > s;
  for (int i = 0; i < (int)m; i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    if (x > y) swap(x, y);
    s.push_back({x, y});
  }
  sort((s).begin(), (s).end());
  for (int i = 2; i < (int)n + 1; i++) {
    if (n % i != 0) continue;
    int t = n / i;
    for (int d : {t}) {
      vector<pair<int, int> > f;
      for (auto p : s) {
        int e1 = modd(p.first + d, n);
        int e2 = modd(p.second + d, n);
        if (e1 > e2) swap(e1, e2);
        f.push_back({e1, e2});
      }
      sort((f).begin(), (f).end());
      if (s == f) {
        cout << "Yes"
             << "\n";
        return;
      }
    }
  }
  {
    cout << "No"
         << "\n";
    return;
  };
}

```