#include <bits/stdc++.h>
using namespace std;
inline long long fpow(long long n, long long k, int p = 998244353) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
inline long long inv(long long a, long long p = 998244353) {
  return fpow(a, p - 2, p);
}
inline long long addmod(long long a, long long val, long long p = 998244353) {
  {
    if ((a = (a + val)) >= p) a -= p;
  }
  return a;
}
inline long long submod(long long a, long long val, long long p = 998244353) {
  {
    if ((a = (a - val)) < 0) a += p;
  }
  return a;
}
inline long long mult(long long a, long long b, long long p = 998244353) {
  return (long long)a * b % p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  set<pair<int, int> > s;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    s.insert(make_pair(x - 1, y - 1));
  }
  vector<int> div;
  for (int i = 1; i < n; i++) {
    if (n % i == 0) div.push_back(i);
  }
  bool check = true;
  for (int i = 0; i < div.size(); i++) {
    check = true;
    for (auto p : s) {
      if (s.find(make_pair((p.first + div[i]) % n, (p.second + div[i]) % n)) ==
              s.end() &&
          s.find(make_pair((p.second + div[i]) % n, (p.first + div[i]) % n)) ==
              s.end()) {
        check = false;
        break;
      }
    }
    if (check == true) break;
  }
  if (check)
    cout << "Yes"
         << "\n";
  else
    cout << "No"
         << "\n";
}

### 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;
inline long long fpow(long long n, long long k, int p = 998244353) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
inline long long inv(long long a, long long p = 998244353) {
  return fpow(a, p - 2, p);
}
inline long long addmod(long long a, long long val, long long p = 998244353) {
  {
    if ((a = (a + val)) >= p) a -= p;
  }
  return a;
}
inline long long submod(long long a, long long val, long long p = 998244353) {
  {
    if ((a = (a - val)) < 0) a += p;
  }
  return a;
}
inline long long mult(long long a, long long b, long long p = 998244353) {
  return (long long)a * b % p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  set<pair<int, int> > s;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    s.insert(make_pair(x - 1, y - 1));
  }
  vector<int> div;
  for (int i = 1; i < n; i++) {
    if (n % i == 0) div.push_back(i);
  }
  bool check = true;
  for (int i = 0; i < div.size(); i++) {
    check = true;
    for (auto p : s) {
      if (s.find(make_pair((p.first + div[i]) % n, (p.second + div[i]) % n)) ==
              s.end() &&
          s.find(make_pair((p.second + div[i]) % n, (p.first + div[i]) % n)) ==
              s.end()) {
        check = false;
        break;
      }
    }
    if (check == true) break;
  }
  if (check)
    cout << "Yes"
         << "\n";
  else
    cout << "No"
         << "\n";
}

```