#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const double PI = acos(-1.0);
const long long int mod = 1e9 + 7;
const int MAXN = 1e6 + 5;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
void cp() {
  int n, m;
  cin >> n >> m;
  map<int, vector<int>> dist;
  for (int i = 0; i < m; i++) {
    int a, b, u, v, d;
    cin >> a >> b;
    u = max(a, b), v = min(a, b);
    d = u - v;
    if (d > n / 2) d = (-d + n) % n;
    dist[d].push_back(u);
    dist[d].push_back(v);
  }
  bool ok = true;
  long long int x = 1;
  for (auto &it : dist) {
    sort(it.second.begin(), it.second.end());
    vector<int> pat;
    for (int i = 0; i < ((int)it.second.size()); i++)
      pat.push_back(
          (it.second[(i + 1) % ((int)it.second.size())] - it.second[i] + n) %
          n);
    vector<int> failure(((int)pat.size()));
    for (int i = 1, k = 0; i < ((int)pat.size()); i++) {
      while (k && pat[i] != pat[k]) k = failure[k - 1];
      if (pat[i] == pat[k]) k++;
      failure[i] = k;
    }
    int rep = failure.back();
    if (rep && ((int)pat.size()) % (((int)pat.size()) - rep) == 0) {
      int len = 0;
      rep = (((int)pat.size()) - rep);
      for (int i = 0; i < rep; i++) len += pat[i];
      x = lcm(x, len);
    } else
      ok = false;
    if (!ok || x >= n) break;
  }
  if (ok && x < n)
    cout << "Yes\n";
  else
    cout << "No\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  t = 1;
  while (t--) {
    cp();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const double eps = 1e-9;
const double PI = acos(-1.0);
const long long int mod = 1e9 + 7;
const int MAXN = 1e6 + 5;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
void cp() {
  int n, m;
  cin >> n >> m;
  map<int, vector<int>> dist;
  for (int i = 0; i < m; i++) {
    int a, b, u, v, d;
    cin >> a >> b;
    u = max(a, b), v = min(a, b);
    d = u - v;
    if (d > n / 2) d = (-d + n) % n;
    dist[d].push_back(u);
    dist[d].push_back(v);
  }
  bool ok = true;
  long long int x = 1;
  for (auto &it : dist) {
    sort(it.second.begin(), it.second.end());
    vector<int> pat;
    for (int i = 0; i < ((int)it.second.size()); i++)
      pat.push_back(
          (it.second[(i + 1) % ((int)it.second.size())] - it.second[i] + n) %
          n);
    vector<int> failure(((int)pat.size()));
    for (int i = 1, k = 0; i < ((int)pat.size()); i++) {
      while (k && pat[i] != pat[k]) k = failure[k - 1];
      if (pat[i] == pat[k]) k++;
      failure[i] = k;
    }
    int rep = failure.back();
    if (rep && ((int)pat.size()) % (((int)pat.size()) - rep) == 0) {
      int len = 0;
      rep = (((int)pat.size()) - rep);
      for (int i = 0; i < rep; i++) len += pat[i];
      x = lcm(x, len);
    } else
      ok = false;
    if (!ok || x >= n) break;
  }
  if (ok && x < n)
    cout << "Yes\n";
  else
    cout << "No\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  t = 1;
  while (t--) {
    cp();
  }
  return 0;
}

```