#include <bits/stdc++.h>
using namespace std;
vector<int> seg[100001];
int n, m;
int dist(int a, int b) {
  int dab = b - a;
  if (dab < 0) dab += n;
  int dba = n - dab;
  return dab < dba ? dab : dba;
}
int main() {
  for (scanf("%d%d", &n, &m); m--;) {
    int a, b;
    scanf("%d%d", &a, &b);
    int d = dist(a, b);
    seg[a].push_back(d);
    seg[b].push_back(d);
  }
  for (int i = 1; i <= n; ++i) sort(seg[i].begin(), seg[i].end());
  for (int i = 1; i < n; ++i) {
    if (n % i) continue;
    int l = 1, r = 1 + i;
    bool flag = 1;
    for (; flag && l <= n; ++l, ++r) {
      if (r > n) r -= n;
      if (seg[l].size() == seg[r].size()) {
        for (int j = 0; flag && j < seg[l].size(); ++j) {
          flag &= seg[l][j] == seg[r][j];
        }
      } else
        flag = 0;
    }
    if (flag) {
      printf("Yes");
      return 0;
    }
  }
  printf("No");
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<int> seg[100001];
int n, m;
int dist(int a, int b) {
  int dab = b - a;
  if (dab < 0) dab += n;
  int dba = n - dab;
  return dab < dba ? dab : dba;
}
int main() {
  for (scanf("%d%d", &n, &m); m--;) {
    int a, b;
    scanf("%d%d", &a, &b);
    int d = dist(a, b);
    seg[a].push_back(d);
    seg[b].push_back(d);
  }
  for (int i = 1; i <= n; ++i) sort(seg[i].begin(), seg[i].end());
  for (int i = 1; i < n; ++i) {
    if (n % i) continue;
    int l = 1, r = 1 + i;
    bool flag = 1;
    for (; flag && l <= n; ++l, ++r) {
      if (r > n) r -= n;
      if (seg[l].size() == seg[r].size()) {
        for (int j = 0; flag && j < seg[l].size(); ++j) {
          flag &= seg[l][j] == seg[r][j];
        }
      } else
        flag = 0;
    }
    if (flag) {
      printf("Yes");
      return 0;
    }
  }
  printf("No");
  return 0;
}

```