#include <bits/stdc++.h>
using namespace std;
long long int p[200008][2];
unordered_set<long long int> se;
vector<long long> v;
long long int hsh(long long int x, long long int y) {
  return x + y * 1000000LL;
}
int main() {
  long long int n, m;
  scanf("%lld%lld", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%lld%lld", &p[i][0], &p[i][1]);
    se.insert(hsh(p[i][0], p[i][1]));
    se.insert(hsh(p[i][1], p[i][0]));
  }
  int flag = 1;
  long long int t = sqrt(n);
  v.push_back(1);
  for (long long i = 2; i <= t; i++) {
    v.push_back(i);
    v.push_back(n / i);
  }
  for (int i = 0; i < v.size() && flag; i++) {
    int flag2 = 1;
    for (int j = 0; j < m && flag2; j++) {
      long long int x = p[j][0];
      long long int y = p[j][1];
      x = (x + v[i]) % n;
      y = (y + v[i]) % n;
      if (x == 0) {
        x = n;
      }
      if (y == 0) {
        y = n;
      }
      if (se.find(hsh(x, y)) == se.end()) {
        flag2 = 0;
      }
    }
    if (flag2) {
      flag = 0;
    }
  }
  if (flag)
    printf("No\n");
  else
    printf("Yes\n");
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
long long int p[200008][2];
unordered_set<long long int> se;
vector<long long> v;
long long int hsh(long long int x, long long int y) {
  return x + y * 1000000LL;
}
int main() {
  long long int n, m;
  scanf("%lld%lld", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%lld%lld", &p[i][0], &p[i][1]);
    se.insert(hsh(p[i][0], p[i][1]));
    se.insert(hsh(p[i][1], p[i][0]));
  }
  int flag = 1;
  long long int t = sqrt(n);
  v.push_back(1);
  for (long long i = 2; i <= t; i++) {
    v.push_back(i);
    v.push_back(n / i);
  }
  for (int i = 0; i < v.size() && flag; i++) {
    int flag2 = 1;
    for (int j = 0; j < m && flag2; j++) {
      long long int x = p[j][0];
      long long int y = p[j][1];
      x = (x + v[i]) % n;
      y = (y + v[i]) % n;
      if (x == 0) {
        x = n;
      }
      if (y == 0) {
        y = n;
      }
      if (se.find(hsh(x, y)) == se.end()) {
        flag2 = 0;
      }
    }
    if (flag2) {
      flag = 0;
    }
  }
  if (flag)
    printf("No\n");
  else
    printf("Yes\n");
  return 0;
}

```