#include <bits/stdc++.h>
using namespace std;
map<pair<int, int>, bool> M;
map<pair<int, int>, bool>::iterator Mit1, Mit2;
bool IsPossible(int N) {
  int i, f, limit = sqrt(N), a, b;
  if (limit * limit > N)
    --limit;
  else if ((limit + 1) * (limit + 1) <= N)
    ++limit;
  for (i = 1; i <= limit; ++i) {
    if (N % i != 0) continue;
    f = i;
    for (Mit1 = M.begin(); Mit1 != M.end(); ++Mit1) {
      a = ((Mit1->first).first + f) % N;
      b = ((Mit1->first).second + f) % N;
      Mit2 = M.find({a, b});
      if (Mit2 == M.end()) break;
    }
    if (Mit1 == M.end()) return true;
    f = N / i;
    if (f == N || f == i) continue;
    for (Mit1 = M.begin(); Mit1 != M.end(); ++Mit1) {
      a = ((Mit1->first).first + f) % N;
      b = ((Mit1->first).second + f) % N;
      Mit2 = M.find({a, b});
      if (Mit2 == M.end()) break;
    }
    if (Mit1 == M.end()) return true;
  }
  return false;
}
int main() {
  int N, m, i, a, b;
  scanf("%d%d", &N, &m);
  for (i = 1; i <= m; ++i) {
    scanf("%d%d", &a, &b);
    M[{a - 1, b - 1}] = true;
    M[{b - 1, a - 1}] = true;
  }
  if (IsPossible(N))
    puts("YES");
  else
    puts("NO");
  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;
map<pair<int, int>, bool> M;
map<pair<int, int>, bool>::iterator Mit1, Mit2;
bool IsPossible(int N) {
  int i, f, limit = sqrt(N), a, b;
  if (limit * limit > N)
    --limit;
  else if ((limit + 1) * (limit + 1) <= N)
    ++limit;
  for (i = 1; i <= limit; ++i) {
    if (N % i != 0) continue;
    f = i;
    for (Mit1 = M.begin(); Mit1 != M.end(); ++Mit1) {
      a = ((Mit1->first).first + f) % N;
      b = ((Mit1->first).second + f) % N;
      Mit2 = M.find({a, b});
      if (Mit2 == M.end()) break;
    }
    if (Mit1 == M.end()) return true;
    f = N / i;
    if (f == N || f == i) continue;
    for (Mit1 = M.begin(); Mit1 != M.end(); ++Mit1) {
      a = ((Mit1->first).first + f) % N;
      b = ((Mit1->first).second + f) % N;
      Mit2 = M.find({a, b});
      if (Mit2 == M.end()) break;
    }
    if (Mit1 == M.end()) return true;
  }
  return false;
}
int main() {
  int N, m, i, a, b;
  scanf("%d%d", &N, &m);
  for (i = 1; i <= m; ++i) {
    scanf("%d%d", &a, &b);
    M[{a - 1, b - 1}] = true;
    M[{b - 1, a - 1}] = true;
  }
  if (IsPossible(N))
    puts("YES");
  else
    puts("NO");
  return 0;
}

```