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>
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 100000;
vector<long long> d1[MAXN], d2[MAXN];
void linear_sort(vector<pair<long long, long long> >& a) {
  for (long long j = 0; j < MAXN; ++j) {
    d1[j].clear();
    d2[j].clear();
  }
  for (long long i = 0; i < a.size(); ++i) d1[a[i].second].emplace_back(i);
  for (long long j = 0; j < MAXN; ++j) {
    for (auto& i : d1[j]) {
      d2[a[i].first].emplace_back(i);
    }
  }
  vector<pair<long long, long long> > b(a.size());
  long long ind = 0;
  for (long long j = 0; j < MAXN; ++j) {
    for (auto& i : d2[j]) {
      b[ind++] = a[i];
    }
  }
  a = b;
}
void Solve() {
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, long long> > a(m);
  for (signed 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);
  }
  linear_sort(a);
  vector<long long> dv = {1};
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      dv.emplace_back(i);
      if (i * i != n) dv.emplace_back(n / i);
    }
  }
  for (auto& d : dv) {
    auto b = a;
    for (auto& p : b) {
      p.first += d;
      if (p.first >= n) p.first -= n;
      p.second += d;
      if (p.second >= n) p.second -= n;
      if (p.first > p.second) swap(p.first, p.second);
    }
    linear_sort(b);
    bool eq = 1;
    for (long long i = 0; eq && i < m; ++i) {
      eq = a[i] == b[i];
    }
    if (eq) return void(cout << "Yes");
  }
  cout << "No";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  Solve();
  return 0;
}
