#include <bits/stdc++.h>
using namespace std;
double const EPS = 1e-9, PI = acos(-1);
const int N = 2e5 + 9, M = 2e3 + 9, OO = 1e9 + 7, MOD = 1e9 + 7;
const long long inf = 1e18;
pair<int, int> pts[N];
int main() {
  cout << fixed << setprecision(9), ios::sync_with_stdio(false),
      cin.tie(nullptr), cout.tie(nullptr);
  int n, m, u, v;
  map<pair<int, int>, bool> mp;
  cin >> n >> m;
  vector<int> fact;
  for (int i = 2; (long long)i * (long long)i <= n; ++i) {
    if (n % i == 0) {
      fact.push_back(i);
      if (n / i != i) fact.push_back(n / i);
    }
  }
  fact.push_back(1);
  sort(fact.begin(), fact.end());
  for (int i = 0; i < m; ++i) {
    cin >> u >> v;
    if (u > v) swap(u, v);
    mp[{u, v}], pts[i] = {u, v};
  }
  for (auto it : fact) {
    int cnt = 0;
    for (int i = 0; i < m; ++i) {
      u = (pts[i].first + it), v = (pts[i].second + it);
      if (u > n) u %= n;
      if (v > n) v %= n;
      if (u > v) swap(u, v);
      if (mp.find({u, v}) != mp.end()) ++cnt;
    }
    if (cnt == m) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
  return 0;
}

### Prompt

Create a solution in cpp for 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;
double const EPS = 1e-9, PI = acos(-1);
const int N = 2e5 + 9, M = 2e3 + 9, OO = 1e9 + 7, MOD = 1e9 + 7;
const long long inf = 1e18;
pair<int, int> pts[N];
int main() {
  cout << fixed << setprecision(9), ios::sync_with_stdio(false),
      cin.tie(nullptr), cout.tie(nullptr);
  int n, m, u, v;
  map<pair<int, int>, bool> mp;
  cin >> n >> m;
  vector<int> fact;
  for (int i = 2; (long long)i * (long long)i <= n; ++i) {
    if (n % i == 0) {
      fact.push_back(i);
      if (n / i != i) fact.push_back(n / i);
    }
  }
  fact.push_back(1);
  sort(fact.begin(), fact.end());
  for (int i = 0; i < m; ++i) {
    cin >> u >> v;
    if (u > v) swap(u, v);
    mp[{u, v}], pts[i] = {u, v};
  }
  for (auto it : fact) {
    int cnt = 0;
    for (int i = 0; i < m; ++i) {
      u = (pts[i].first + it), v = (pts[i].second + it);
      if (u > n) u %= n;
      if (v > n) v %= n;
      if (u > v) swap(u, v);
      if (mp.find({u, v}) != mp.end()) ++cnt;
    }
    if (cnt == m) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
  return 0;
}

```