#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.1415926535897;
const long long mod = 1000000007LL;
const long long INF = 1e18;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<vector<long long>> g(n + 1, vector<long long>());
  long long u, v;
  for (long long i = 0; i < m; i++) {
    cin >> u >> v;
    g[u].push_back(v - u);
    if (g[u].back() < 0) g[u].back() += n;
    g[v].push_back(u - v);
    if (g[v].back() < 0) g[v].back() += n;
  }
  for (auto &v : g) {
    sort(v.begin(), v.end());
  }
  auto f = [&](long long x) { return (x % n == 0 ? n : x % n); };
  auto ch = [&](long long i) {
    bool fl = 1;
    for (long long j = 1; j <= n; j++) {
      fl = fl && (g[j] == g[f(j + i)]);
    }
    return fl;
  };
  for (long long i = 1; i * i <= n; i++) {
    if (n % i != 0) continue;
    if (ch(i) || (n / i != n && ch(n / i))) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
  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 long double PI = 3.1415926535897;
const long long mod = 1000000007LL;
const long long INF = 1e18;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<vector<long long>> g(n + 1, vector<long long>());
  long long u, v;
  for (long long i = 0; i < m; i++) {
    cin >> u >> v;
    g[u].push_back(v - u);
    if (g[u].back() < 0) g[u].back() += n;
    g[v].push_back(u - v);
    if (g[v].back() < 0) g[v].back() += n;
  }
  for (auto &v : g) {
    sort(v.begin(), v.end());
  }
  auto f = [&](long long x) { return (x % n == 0 ? n : x % n); };
  auto ch = [&](long long i) {
    bool fl = 1;
    for (long long j = 1; j <= n; j++) {
      fl = fl && (g[j] == g[f(j + i)]);
    }
    return fl;
  };
  for (long long i = 1; i * i <= n; i++) {
    if (n % i != 0) continue;
    if (ch(i) || (n / i != n && ch(n / i))) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
  return 0;
}

```