#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
vector<int> g[N];
int a[N], b[N], n, m;
int dis(int u, int v) {
  if (v > u) swap(v, u);
  return min(u - v, n - (u - v));
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  bool flag = false;
  for (int i = 1; i < n; i++) {
    if (flag) break;
    if (n % i != 0) continue;
    bool fl = true;
    for (int j = 0; j < n; j++) {
      int u = j, v = (j + i) % n;
      if (g[u].size() != g[v].size()) {
        fl = false;
        break;
      }
      int num = g[u].size();
      for (int k = 0; k < num; k++) {
        a[k] = dis(u, g[u][k]);
        b[k] = dis(v, g[v][k]);
      }
      sort(a, a + num);
      sort(b, b + num);
      for (int k = 0; k < num; k++)
        if (a[k] != b[k]) fl = false;
    }
    if (fl) flag = true;
  }
  if (flag)
    puts("Yes");
  else
    puts("No");
  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;
const int N = 2e5 + 5;
vector<int> g[N];
int a[N], b[N], n, m;
int dis(int u, int v) {
  if (v > u) swap(v, u);
  return min(u - v, n - (u - v));
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  bool flag = false;
  for (int i = 1; i < n; i++) {
    if (flag) break;
    if (n % i != 0) continue;
    bool fl = true;
    for (int j = 0; j < n; j++) {
      int u = j, v = (j + i) % n;
      if (g[u].size() != g[v].size()) {
        fl = false;
        break;
      }
      int num = g[u].size();
      for (int k = 0; k < num; k++) {
        a[k] = dis(u, g[u][k]);
        b[k] = dis(v, g[v][k]);
      }
      sort(a, a + num);
      sort(b, b + num);
      for (int k = 0; k < num; k++)
        if (a[k] != b[k]) fl = false;
    }
    if (fl) flag = true;
  }
  if (flag)
    puts("Yes");
  else
    puts("No");
  return 0;
}

```