#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " , ";
  __f(comma + 1, args...);
}
int n, m;
vector<pair<int, int> > a;
bool check(int k) {
  int i;
  vector<pair<int, int> > c(m);
  for (i = 0; i < m; ++i) {
    c[i].first = (a[i].first + k) % n;
    c[i].second = (a[i].second + k) % n;
    if (c[i].first > c[i].second) swap(c[i].first, c[i].second);
  }
  sort(c.begin(), c.end());
  return c == a;
}
int main() {
  ios::sync_with_stdio(0);
  int i, j, k;
  cin >> n >> m;
  a.resize(m);
  for (i = 0; i < m; ++i) {
    cin >> a[i].first >> a[i].second;
    a[i].first--;
    a[i].second--;
    if (a[i].first > a[i].second) swap(a[i].first, a[i].second);
  }
  sort(a.begin(), a.end());
  bool can = false;
  for (i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      can = can || check(i);
      if (n / i != n) can = can || check(n / i);
    }
  }
  cout << (can ? "Yes\n" : "No\n");
  return 0;
}

### Prompt

Generate a CPP solution 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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " , ";
  __f(comma + 1, args...);
}
int n, m;
vector<pair<int, int> > a;
bool check(int k) {
  int i;
  vector<pair<int, int> > c(m);
  for (i = 0; i < m; ++i) {
    c[i].first = (a[i].first + k) % n;
    c[i].second = (a[i].second + k) % n;
    if (c[i].first > c[i].second) swap(c[i].first, c[i].second);
  }
  sort(c.begin(), c.end());
  return c == a;
}
int main() {
  ios::sync_with_stdio(0);
  int i, j, k;
  cin >> n >> m;
  a.resize(m);
  for (i = 0; i < m; ++i) {
    cin >> a[i].first >> a[i].second;
    a[i].first--;
    a[i].second--;
    if (a[i].first > a[i].second) swap(a[i].first, a[i].second);
  }
  sort(a.begin(), a.end());
  bool can = false;
  for (i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      can = can || check(i);
      if (n / i != n) can = can || check(n / i);
    }
  }
  cout << (can ? "Yes\n" : "No\n");
  return 0;
}

```