#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using vi = vector<int>;
using ll = long long;
const int N = 1e5;
vi v[N];
vi pre;
void precompute(const string& pat) {
  int m = int((pat).size()), p = -1;
  pre = vi(m + 1);
  pre[0] = -1;
  for (int i = int(1); i < int(m + 1); i++) {
    while (p != -1 && pat[p] != pat[i - 1]) p = pre[p];
    p++;
    pre[i] = p;
  }
}
vi match(const string& s, const string& pat) {
  vi res;
  int n = int((s).size()), m = int((pat).size()), p = 0;
  for (int i = int(0); i < int(n); i++) {
    while (p != -1 && pat[p] != s[i]) p = pre[p];
    p++;
    if (p == m) {
      res.emplace_back(i - m + 1);
      p = pre[p];
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  while (m--) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    if (a > b) swap(a, b);
    v[a].emplace_back(b);
    v[b].emplace_back(n + a);
  }
  string p;
  for (int a = int(0); a < int(n); a++) {
    sort(v[a].begin(), v[a].end());
    for (int b : v[a]) p += to_string(b - a);
    p += to_string(-1);
  }
  string s = p + p;
  precompute(p);
  cout << (int((match(s, p)).size()) > 2 ? "Yes" : "No") << '\n';
  return 0;
}

### Prompt

Generate 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;
using pii = pair<int, int>;
using vi = vector<int>;
using ll = long long;
const int N = 1e5;
vi v[N];
vi pre;
void precompute(const string& pat) {
  int m = int((pat).size()), p = -1;
  pre = vi(m + 1);
  pre[0] = -1;
  for (int i = int(1); i < int(m + 1); i++) {
    while (p != -1 && pat[p] != pat[i - 1]) p = pre[p];
    p++;
    pre[i] = p;
  }
}
vi match(const string& s, const string& pat) {
  vi res;
  int n = int((s).size()), m = int((pat).size()), p = 0;
  for (int i = int(0); i < int(n); i++) {
    while (p != -1 && pat[p] != s[i]) p = pre[p];
    p++;
    if (p == m) {
      res.emplace_back(i - m + 1);
      p = pre[p];
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  while (m--) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    if (a > b) swap(a, b);
    v[a].emplace_back(b);
    v[b].emplace_back(n + a);
  }
  string p;
  for (int a = int(0); a < int(n); a++) {
    sort(v[a].begin(), v[a].end());
    for (int b : v[a]) p += to_string(b - a);
    p += to_string(-1);
  }
  string s = p + p;
  precompute(p);
  cout << (int((match(s, p)).size()) > 2 ? "Yes" : "No") << '\n';
  return 0;
}

```