#include <bits/stdc++.h>
using namespace std;
using int64 = long long;
void DBG() { cerr << "]" << '\n'; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << h;
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > a(m);
  for (auto &[x, y] : a) {
    cin >> x >> y;
    x--, y--;
    if (x > y) swap(x, y);
  }
  vector<int> ds;
  for (int i = 1; i * i <= n; i++) {
    if (n % i) continue;
    ds.push_back(i);
    if (i != n / i) ds.push_back(n / i);
  }
  sort(ds.begin(), ds.end());
  ds.pop_back();
  vector<pair<int, int> > q(m), w(m);
  for (int d : ds) {
    int i = 0;
    for (auto [x, y] : a) {
      q[i] = {x, y};
      int nx = (x + d) % n;
      int ny = (y + d) % n;
      if (nx > ny) swap(nx, ny);
      w[i] = {nx, ny};
      i++;
    }
    sort(q.begin(), q.end());
    sort(w.begin(), w.end());
    int bad = 0;
    for (int i = 0; i < m; i++) {
      if (q[i] != w[i]) {
        bad = 1;
        break;
      }
    }
    if (!bad) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No\n";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
using int64 = long long;
void DBG() { cerr << "]" << '\n'; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << h;
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > a(m);
  for (auto &[x, y] : a) {
    cin >> x >> y;
    x--, y--;
    if (x > y) swap(x, y);
  }
  vector<int> ds;
  for (int i = 1; i * i <= n; i++) {
    if (n % i) continue;
    ds.push_back(i);
    if (i != n / i) ds.push_back(n / i);
  }
  sort(ds.begin(), ds.end());
  ds.pop_back();
  vector<pair<int, int> > q(m), w(m);
  for (int d : ds) {
    int i = 0;
    for (auto [x, y] : a) {
      q[i] = {x, y};
      int nx = (x + d) % n;
      int ny = (y + d) % n;
      if (nx > ny) swap(nx, ny);
      w[i] = {nx, ny};
      i++;
    }
    sort(q.begin(), q.end());
    sort(w.begin(), w.end());
    int bad = 0;
    for (int i = 0; i < m; i++) {
      if (q[i] != w[i]) {
        bad = 1;
        break;
      }
    }
    if (!bad) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No\n";
  return 0;
}

```