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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
constexpr int INF = (1 << 30);
constexpr ll INFL = (1ll << 60);
constexpr ll MOD = 1000000007;
bool solve(vector<pair<int, int>> ab, int k, int n) {
  auto next = ab;
  for (auto& p : next) {
    p.first = (p.first + k) % n;
    p.second = (p.second + k) % n;
    if (p.first > p.second) swap(p.first, p.second);
  }
  sort(next.begin(), next.end());
  for (ll((i)) = (0); ((i)) < ((ab.size())); ((i))++)
    if (ab[i] != next[i]) return false;
  return true;
}
int main() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> ab(m);
  for (ll((i)) = (0); ((i)) < ((m)); ((i))++) {
    cin >> ab[i].first >> ab[i].second;
    ab[i].first--;
    ab[i].second--;
    if (ab[i].first > ab[i].second) swap(ab[i].first, ab[i].second);
  }
  sort(ab.begin(), ab.end());
  set<int> cands;
  for (ll(i) = (1); (i) < (n + 1); (i)++)
    if (n % i == 0) {
      cands.insert(i);
      cands.insert(n / i);
    }
  if (cands.find(n) != cands.end()) cands.erase(n);
  for (auto num : cands) {
    if (solve(ab, num, n)) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
  return 0;
}
