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;
struct hasher {
  inline size_t operator()(const pair<int, int>& v) const {
    return v.first * 31 + v.second;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<int> d;
  d.push_back(1);
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      d.push_back(i);
      if (i != n / i) d.push_back(n / i);
    }
  }
  unordered_set<pair<long long, long long>, hasher> ms;
  while (m--) {
    int a, b;
    cin >> a >> b;
    ms.insert({a, b});
    ms.insert({b, a});
  }
  for (int x : d) {
    int cnt = n / x;
    if (ms.size() % cnt != 0) continue;
    auto kek = ms;
    bool ok = true;
    while (!kek.empty()) {
      auto cur = *kek.begin();
      auto start = cur;
      kek.erase(kek.begin());
      while (1) {
        cur.first += x;
        cur.second += x;
        if (cur.second > n) cur.second -= n;
        if (cur.first > n) cur.first -= n;
        if (cur.first == start.first) break;
        auto it = kek.find(cur);
        if (it == kek.end()) {
          ok = false;
          break;
        }
        kek.erase(it);
      }
      if (!ok) break;
    }
    if (ok) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No\n";
  return 0;
}
