#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using piii = pair<int, pii>;
using plll = pair<ll, pll>;
using pib = pair<int, bool>;
using pdi = pair<double, int>;
using pid = pair<int, double>;
using ld = long double;
using piiii = pair<pii, pii>;
int n, m, arr[200000][2];
map<pii, bool> mp;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> arr[i][0] >> arr[i][1];
    if (arr[i][1] > arr[i][0]) swap(arr[i][0], arr[i][1]);
    mp[{arr[i][0], arr[i][1]}] = true;
  }
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      bool first = true;
      for (int j = 0; j < m; j++) {
        int a = arr[j][0] + i;
        int b = arr[j][1] + i;
        if (a > n) a -= n;
        if (b > n) b -= n;
        if (b > a) swap(a, b);
        if (!mp.count({a, b})) {
          first = false;
          break;
        }
      }
      if (first) {
        cout << "Yes"
             << "\n";
        return 0;
      }
    }
  }
  cout << "No"
       << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using piii = pair<int, pii>;
using plll = pair<ll, pll>;
using pib = pair<int, bool>;
using pdi = pair<double, int>;
using pid = pair<int, double>;
using ld = long double;
using piiii = pair<pii, pii>;
int n, m, arr[200000][2];
map<pii, bool> mp;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> arr[i][0] >> arr[i][1];
    if (arr[i][1] > arr[i][0]) swap(arr[i][0], arr[i][1]);
    mp[{arr[i][0], arr[i][1]}] = true;
  }
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      bool first = true;
      for (int j = 0; j < m; j++) {
        int a = arr[j][0] + i;
        int b = arr[j][1] + i;
        if (a > n) a -= n;
        if (b > n) b -= n;
        if (b > a) swap(a, b);
        if (!mp.count({a, b})) {
          first = false;
          break;
        }
      }
      if (first) {
        cout << "Yes"
             << "\n";
        return 0;
      }
    }
  }
  cout << "No"
       << "\n";
  return 0;
}

```