#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int dx[8] = {1, -1, 0, 0, -1, -1, 1, 1};
int dy[8] = {0, 0, -1, 1, -1, 1, -1, 1};
vector<long long> par(300010, -1), vis(300010, 0), dis(300010, -1);
vector<pair<long long, long long>> adj[300010];
int myrandom(int i) { return std::rand() % i; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  set<pair<int, int>> st;
  for (long long i = 0; i < k; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b) {
      swap(a, b);
    }
    st.insert({a, b});
  }
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      bool ok = true;
      for (auto j : st) {
        int x = j.first + i;
        int y = j.second + i;
        if (x > n) {
          x = x - n;
        }
        if (y > n) {
          y = y - n;
        }
        if (x > y) {
          swap(x, y);
        }
        if (st.find({x, y}) == st.end()) {
          ok = false;
          break;
        }
      }
      if (ok) {
        cout << "Yes" << '\n';
        return 0;
      }
    }
  }
  cout << "No" << '\n';
}

### Prompt

Please create a solution in CPP 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int dx[8] = {1, -1, 0, 0, -1, -1, 1, 1};
int dy[8] = {0, 0, -1, 1, -1, 1, -1, 1};
vector<long long> par(300010, -1), vis(300010, 0), dis(300010, -1);
vector<pair<long long, long long>> adj[300010];
int myrandom(int i) { return std::rand() % i; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  set<pair<int, int>> st;
  for (long long i = 0; i < k; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b) {
      swap(a, b);
    }
    st.insert({a, b});
  }
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      bool ok = true;
      for (auto j : st) {
        int x = j.first + i;
        int y = j.second + i;
        if (x > n) {
          x = x - n;
        }
        if (y > n) {
          y = y - n;
        }
        if (x > y) {
          swap(x, y);
        }
        if (st.find({x, y}) == st.end()) {
          ok = false;
          break;
        }
      }
      if (ok) {
        cout << "Yes" << '\n';
        return 0;
      }
    }
  }
  cout << "No" << '\n';
}

```