#include <bits/stdc++.h>
using namespace std;
int n, m;
vector<int> divs;
pair<int, int> segs[200010];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      divs.push_back(i);
      if (i != 1) divs.push_back(n / i);
    }
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &segs[i].first, &segs[i].second);
    segs[i].first %= n;
    segs[i].second %= n;
    if (segs[i].first > segs[i].second) swap(segs[i].first, segs[i].second);
  }
  sort(segs, segs + m);
  int s = (int)divs.size();
  for (int i = 0; i < s; i++) {
    pair<int, int> new_segs[200010];
    for (int j = 0; j < m; j++) {
      new_segs[j] = make_pair((segs[j].first + divs[i]) % n,
                              (segs[j].second + divs[i]) % n);
      if (new_segs[j].first > new_segs[j].second)
        swap(new_segs[j].first, new_segs[j].second);
    }
    sort(new_segs, new_segs + m);
    bool poss = true;
    for (int j = 0; j < m; j++) {
      if (new_segs[j] != segs[j]) {
        poss = false;
        break;
      }
    }
    if (poss) {
      printf("Yes\n");
      return 0;
    }
    for (int j = 0; j < m; j++) {
      new_segs[j] = make_pair((segs[j].first + n - divs[i]) % n,
                              (segs[j].second + n - divs[i]) % n);
      if (new_segs[j].first > new_segs[j].second)
        swap(new_segs[j].first, new_segs[j].second);
    }
    sort(new_segs, new_segs + m);
    poss = true;
    for (int j = 0; j < m; j++) {
      if (new_segs[j] != segs[j]) {
        poss = false;
        break;
      }
    }
    if (poss) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

### 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;
int n, m;
vector<int> divs;
pair<int, int> segs[200010];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      divs.push_back(i);
      if (i != 1) divs.push_back(n / i);
    }
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &segs[i].first, &segs[i].second);
    segs[i].first %= n;
    segs[i].second %= n;
    if (segs[i].first > segs[i].second) swap(segs[i].first, segs[i].second);
  }
  sort(segs, segs + m);
  int s = (int)divs.size();
  for (int i = 0; i < s; i++) {
    pair<int, int> new_segs[200010];
    for (int j = 0; j < m; j++) {
      new_segs[j] = make_pair((segs[j].first + divs[i]) % n,
                              (segs[j].second + divs[i]) % n);
      if (new_segs[j].first > new_segs[j].second)
        swap(new_segs[j].first, new_segs[j].second);
    }
    sort(new_segs, new_segs + m);
    bool poss = true;
    for (int j = 0; j < m; j++) {
      if (new_segs[j] != segs[j]) {
        poss = false;
        break;
      }
    }
    if (poss) {
      printf("Yes\n");
      return 0;
    }
    for (int j = 0; j < m; j++) {
      new_segs[j] = make_pair((segs[j].first + n - divs[i]) % n,
                              (segs[j].second + n - divs[i]) % n);
      if (new_segs[j].first > new_segs[j].second)
        swap(new_segs[j].first, new_segs[j].second);
    }
    sort(new_segs, new_segs + m);
    poss = true;
    for (int j = 0; j < m; j++) {
      if (new_segs[j] != segs[j]) {
        poss = false;
        break;
      }
    }
    if (poss) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

```