#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<int> > v(n + 1);
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    if (b - a == n - b + a) {
      v[a].push_back(b - a);
      v[b].push_back(b - a);
    } else if (b - a < n - b + a)
      v[a].push_back(b - a);
    else
      v[b].push_back(n - b + a);
  }
  for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
  for (int k = 1; k < n; k++) {
    if (n % k) continue;
    bool ok = true;
    for (int i = 1; i <= n; i++)
      if (v[i] != v[(i + k > n) ? (i + k - n) : (i + k)]) ok = false;
    if (ok) return printf("Yes\n"), 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 N = 2e5 + 5;
int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<int> > v(n + 1);
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    if (b - a == n - b + a) {
      v[a].push_back(b - a);
      v[b].push_back(b - a);
    } else if (b - a < n - b + a)
      v[a].push_back(b - a);
    else
      v[b].push_back(n - b + a);
  }
  for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
  for (int k = 1; k < n; k++) {
    if (n % k) continue;
    bool ok = true;
    for (int i = 1; i <= n; i++)
      if (v[i] != v[(i + k > n) ? (i + k - n) : (i + k)]) ok = false;
    if (ok) return printf("Yes\n"), 0;
  }
  printf("No\n");
}

```