#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
void readv(vector<T> &v, int x) {
  v = vector<T>(x);
  for (int(i) = int(0); i < int(x); ++i) cin >> (v[i]);
}
int n, m;
vector<pair<int, int> > ab;
void _input() {
  cin >> n >> m;
  ab = vector<pair<int, int> >(m);
  for (int(i) = int(0); i < int(m); ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    if (a > b) swap(a, b);
    ab[i] = {a, b};
  }
}
bool solve() { return true; }
const bool DEB = 0;
vector<int> getd(int x) {
  vector<int> ret;
  for (int i = 1; i < x / 2 + 1; ++i) {
    if (x % i == 0) {
      ret.push_back(i);
    }
  }
  return ret;
}
void _solve() {
  vector<int> div = getd(n);
  bool ye = false;
  sort(ab.begin(), ab.end());
  vector<pair<int, int> > vp(m);
  for (auto d : div) {
    for (int(i) = int(0); i < int(m); ++i) {
      vp[i] = {(ab[i].first + d) % n, (ab[i].second + d) % n};
      if (vp[i].first > vp[i].second) swap(vp[i].first, vp[i].second);
    }
    sort(vp.begin(), vp.end());
    ye = vp == ab;
    if (ye) break;
  }
  if (ye)
    cout << "Yes\n";
  else
    cout << "No\n";
}
int main() {
  double st = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int qq = 1;
  int cc = 0;
  while (cc++ < qq) {
    _input();
    _solve();
  }
  cerr << 1000.0 * (1.0 * clock() - st) / (1.0 * CLOCKS_PER_SEC)
       << " ms ellapsed\n";
}

### 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
void readv(vector<T> &v, int x) {
  v = vector<T>(x);
  for (int(i) = int(0); i < int(x); ++i) cin >> (v[i]);
}
int n, m;
vector<pair<int, int> > ab;
void _input() {
  cin >> n >> m;
  ab = vector<pair<int, int> >(m);
  for (int(i) = int(0); i < int(m); ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    if (a > b) swap(a, b);
    ab[i] = {a, b};
  }
}
bool solve() { return true; }
const bool DEB = 0;
vector<int> getd(int x) {
  vector<int> ret;
  for (int i = 1; i < x / 2 + 1; ++i) {
    if (x % i == 0) {
      ret.push_back(i);
    }
  }
  return ret;
}
void _solve() {
  vector<int> div = getd(n);
  bool ye = false;
  sort(ab.begin(), ab.end());
  vector<pair<int, int> > vp(m);
  for (auto d : div) {
    for (int(i) = int(0); i < int(m); ++i) {
      vp[i] = {(ab[i].first + d) % n, (ab[i].second + d) % n};
      if (vp[i].first > vp[i].second) swap(vp[i].first, vp[i].second);
    }
    sort(vp.begin(), vp.end());
    ye = vp == ab;
    if (ye) break;
  }
  if (ye)
    cout << "Yes\n";
  else
    cout << "No\n";
}
int main() {
  double st = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int qq = 1;
  int cc = 0;
  while (cc++ < qq) {
    _input();
    _solve();
  }
  cerr << 1000.0 * (1.0 * clock() - st) / (1.0 * CLOCKS_PER_SEC)
       << " ms ellapsed\n";
}

```