#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)5e5 + 100;
const long long inf = (1ll << 60);
const int mod2 = (int)998244353;
const int mod = (int)1e9 + 7;
string s;
int n;
pair<int, int> a[maxn];
pair<int, int> b[maxn];
int m;
int cnt[maxn];
vector<int> g[maxn];
vector<int> gr[maxn];
int T;
void solve() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a[i * 2].first >> a[i * 2].second;
    a[i * 2].first--;
    a[i * 2].second--;
    a[i * 2 + 1] = a[i * 2];
    swap(a[i * 2 + 1].first, a[i * 2 + 1].second);
  }
  m *= 2;
  sort(a, a + m);
  for (int i = 0; i < m; i++) {
    g[a[i].first].push_back(a[i].second);
  }
  for (int i = 0; i < n; i++) sort(g[i].begin(), g[i].end());
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      for (int j = 0; j < n; j++) {
        gr[j].clear();
      }
      for (int j = 0; j < m; j++) {
        b[j].first = (a[j].first + i) % n;
        b[j].second = (a[j].second + i) % n;
        gr[b[j].first].push_back(b[j].second);
      }
      int ok = 1;
      for (int j = 0; ok && j < n; j++) {
        if (gr[j].size() != g[j].size()) {
          ok = 0;
        } else {
          T++;
          for (int x : g[j]) cnt[x]++;
          for (int x : gr[j]) {
            if (cnt[x] > 0) {
              cnt[x]--;
            } else {
              ok = 0;
            }
          }
          for (int x : g[j]) cnt[x] = 0;
        }
      }
      if (ok) {
        cout << "Yes\n";
        return;
      }
    }
  }
  cout << "No\n";
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### 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;
const int maxn = (int)5e5 + 100;
const long long inf = (1ll << 60);
const int mod2 = (int)998244353;
const int mod = (int)1e9 + 7;
string s;
int n;
pair<int, int> a[maxn];
pair<int, int> b[maxn];
int m;
int cnt[maxn];
vector<int> g[maxn];
vector<int> gr[maxn];
int T;
void solve() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a[i * 2].first >> a[i * 2].second;
    a[i * 2].first--;
    a[i * 2].second--;
    a[i * 2 + 1] = a[i * 2];
    swap(a[i * 2 + 1].first, a[i * 2 + 1].second);
  }
  m *= 2;
  sort(a, a + m);
  for (int i = 0; i < m; i++) {
    g[a[i].first].push_back(a[i].second);
  }
  for (int i = 0; i < n; i++) sort(g[i].begin(), g[i].end());
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      for (int j = 0; j < n; j++) {
        gr[j].clear();
      }
      for (int j = 0; j < m; j++) {
        b[j].first = (a[j].first + i) % n;
        b[j].second = (a[j].second + i) % n;
        gr[b[j].first].push_back(b[j].second);
      }
      int ok = 1;
      for (int j = 0; ok && j < n; j++) {
        if (gr[j].size() != g[j].size()) {
          ok = 0;
        } else {
          T++;
          for (int x : g[j]) cnt[x]++;
          for (int x : gr[j]) {
            if (cnt[x] > 0) {
              cnt[x]--;
            } else {
              ok = 0;
            }
          }
          for (int x : g[j]) cnt[x] = 0;
        }
      }
      if (ok) {
        cout << "Yes\n";
        return;
      }
    }
  }
  cout << "No\n";
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```