#include <bits/stdc++.h>
using namespace std;
const int INF = 3e18;
const int MAXN = 2e5 + 228;
int n, m;
set<pair<int, int>> beg;
int nxt(int ind) {
  if (ind >= n) ind -= n;
  if (ind == 0) return n;
  return ind;
}
bool check(vector<int>& x, vector<int>& y, int add) {
  for (int i = 0; i < m; ++i) {
    int mn2 = nxt(x[i] + add), mx2 = nxt(y[i] + add);
    if (mn2 > mx2) swap(mn2, mx2);
    if (!beg.count({mn2, mx2})) {
      return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  vector<int> x(m), y(m);
  for (int i = 0; i < m; ++i) {
    cin >> x[i] >> y[i];
    if (x[i] > y[i]) {
      swap(x[i], y[i]);
    }
    beg.insert({x[i], y[i]});
  }
  for (int i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      if (check(x, y, i)) {
        cout << "Yes\n";
        return 0;
      }
      if (i != 1) {
        if (check(x, y, n / i)) {
          cout << "Yes\n";
          return 0;
        }
      }
    }
  }
  cout << "No\n";
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
const int INF = 3e18;
const int MAXN = 2e5 + 228;
int n, m;
set<pair<int, int>> beg;
int nxt(int ind) {
  if (ind >= n) ind -= n;
  if (ind == 0) return n;
  return ind;
}
bool check(vector<int>& x, vector<int>& y, int add) {
  for (int i = 0; i < m; ++i) {
    int mn2 = nxt(x[i] + add), mx2 = nxt(y[i] + add);
    if (mn2 > mx2) swap(mn2, mx2);
    if (!beg.count({mn2, mx2})) {
      return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  vector<int> x(m), y(m);
  for (int i = 0; i < m; ++i) {
    cin >> x[i] >> y[i];
    if (x[i] > y[i]) {
      swap(x[i], y[i]);
    }
    beg.insert({x[i], y[i]});
  }
  for (int i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      if (check(x, y, i)) {
        cout << "Yes\n";
        return 0;
      }
      if (i != 1) {
        if (check(x, y, n / i)) {
          cout << "Yes\n";
          return 0;
        }
      }
    }
  }
  cout << "No\n";
  return 0;
}

```