#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 7;
vector<int> fac;
vector<int> vex[MAXN];
int n, m;
void split(int n) {
  int cnt = sqrt(n) + 1;
  fac.push_back(1);
  for (int i = 2; i < cnt; i++) {
    if (n % i == 0) fac.push_back(i), fac.push_back(n / i);
  }
  sort(fac.rbegin(), fac.rend());
}
bool check(int d) {
  int ops = 1;
  int cnt = n / d;
  for (int i = 1; i < cnt; i++) {
    for (int j = 0; j < d; j++) {
      if (vex[ops + j].size() != vex[ops + j + i * d].size()) return false;
      for (int k = 0; k < vex[ops + j].size(); k++) {
        if (vex[ops + j][k] != vex[ops + j + i * d][k]) return false;
      }
    }
  }
  return true;
}
int main() {
  cin >> n >> m;
  int begin, end;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &begin, &end);
    vex[begin].push_back(((end - begin) % n + n) % n);
    vex[end].push_back(((begin - end) % n + n) % n);
  }
  for (int i = 1; i <= n; i++) {
    sort(vex[i].begin(), vex[i].end());
  }
  split(n);
  for (int i = 0; i < fac.size(); i++) {
    int guess = fac[i];
    if (check(guess)) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
}

### 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;
const int MAXN = 1e5 + 7;
vector<int> fac;
vector<int> vex[MAXN];
int n, m;
void split(int n) {
  int cnt = sqrt(n) + 1;
  fac.push_back(1);
  for (int i = 2; i < cnt; i++) {
    if (n % i == 0) fac.push_back(i), fac.push_back(n / i);
  }
  sort(fac.rbegin(), fac.rend());
}
bool check(int d) {
  int ops = 1;
  int cnt = n / d;
  for (int i = 1; i < cnt; i++) {
    for (int j = 0; j < d; j++) {
      if (vex[ops + j].size() != vex[ops + j + i * d].size()) return false;
      for (int k = 0; k < vex[ops + j].size(); k++) {
        if (vex[ops + j][k] != vex[ops + j + i * d][k]) return false;
      }
    }
  }
  return true;
}
int main() {
  cin >> n >> m;
  int begin, end;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &begin, &end);
    vex[begin].push_back(((end - begin) % n + n) % n);
    vex[end].push_back(((begin - end) % n + n) % n);
  }
  for (int i = 1; i <= n; i++) {
    sort(vex[i].begin(), vex[i].end());
  }
  split(n);
  for (int i = 0; i < fac.size(); i++) {
    int guess = fac[i];
    if (check(guess)) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
}

```