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;
int n, m;
set<pair<int, int>> seg;
void test(int k) {
  if (k == n) return;
  for (auto it = seg.begin(); it != seg.end(); ++it) {
    int a = (it->first + k) % n;
    int b = (it->second + k) % n;
    if (seg.find({min(a, b), max(a, b)}) == seg.end()) return;
  }
  cout << "Yes" << endl;
  exit(0);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0, a, b; i < m; ++i) {
    cin >> a >> b;
    --a;
    --b;
    seg.insert({min(a, b), max(a, b)});
  }
  vector<int> divisors;
  for (int i = 1; i * i <= n; ++i) {
    if (n % i) continue;
    divisors.push_back(i);
    if (i * i != n) divisors.push_back(n / i);
  }
  for (auto f : divisors) test(f);
  cout << "No\n";
}
