#include <bits/stdc++.h>
using namespace std;
bool shift(set<pair<int, int>> se, int p, int n) {
  set<pair<int, int>> new_s;
  for (auto it = se.begin(); it != se.end(); it++) {
    pair<int, int> pr = *it;
    int a = (pr.first + p) % n;
    int b = (pr.second + p) % n;
    if (a > b) swap(a, b);
    new_s.insert(make_pair(a, b));
  }
  return (new_s == se);
}
void swap(int &a, int &b) {
  int temp = a;
  a = b;
  b = temp;
}
void test() {
  int n, m;
  cin >> n >> m;
  set<pair<int, int>> uns;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    if (a > b) swap(a, b);
    uns.insert(make_pair(a, b));
  }
  vector<int> primes;
  int x = n;
  int prev = -1;
  for (int p = 2; p <= x; p++) {
    while (x % p == 0) {
      if (prev != p) {
        primes.push_back(p);
        prev = p;
      }
      x /= p;
    }
  }
  bool ans = false;
  for (int p : primes) {
    if (shift(uns, n / p, n)) {
      ans = true;
      break;
    }
  }
  if (ans)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  test();
}

### Prompt

Your task is to create 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;
bool shift(set<pair<int, int>> se, int p, int n) {
  set<pair<int, int>> new_s;
  for (auto it = se.begin(); it != se.end(); it++) {
    pair<int, int> pr = *it;
    int a = (pr.first + p) % n;
    int b = (pr.second + p) % n;
    if (a > b) swap(a, b);
    new_s.insert(make_pair(a, b));
  }
  return (new_s == se);
}
void swap(int &a, int &b) {
  int temp = a;
  a = b;
  b = temp;
}
void test() {
  int n, m;
  cin >> n >> m;
  set<pair<int, int>> uns;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    if (a > b) swap(a, b);
    uns.insert(make_pair(a, b));
  }
  vector<int> primes;
  int x = n;
  int prev = -1;
  for (int p = 2; p <= x; p++) {
    while (x % p == 0) {
      if (prev != p) {
        primes.push_back(p);
        prev = p;
      }
      x /= p;
    }
  }
  bool ans = false;
  for (int p : primes) {
    if (shift(uns, n / p, n)) {
      ans = true;
      break;
    }
  }
  if (ans)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  test();
}

```