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 int inf = 0x3f3f3f3f;
const double eps = 1e-6;
const int mod = 1000000007;
int modpow(int x, int y, int md = mod) {
  if (y == 0) return 1;
  int ret = modpow(x, y >> 1, md);
  ret = (long long)ret * ret % md;
  if (y & 1) ret = (long long)ret * x % md;
  return ret;
}
int n, m;
vector<pair<int, int> > v;
void solve(int x) {
  vector<pair<int, int> > nv;
  for (auto p : v) {
    int a = p.first + x, b = p.second + x;
    if (a > n) a -= n;
    if (b > n) b -= n;
    nv.push_back(make_pair(min(a, b), max(a, b)));
  }
  sort(nv.begin(), nv.end());
  if (nv == v) {
    puts("yES");
    exit(0);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  while (m--) {
    int a, b;
    cin >> a >> b;
    v.push_back(make_pair(min(a, b), max(a, b)));
  }
  sort(v.begin(), v.end());
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      solve(i);
      if (i != 1) solve(n / i);
    }
  }
  puts("nO");
  return 0;
}
