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 pii = pair<int, int>;
using vi = vector<int>;
using ll = long long;
bool hasPeriod(string &s) {
  int n = int((s).size());
  vi f(n);
  for (int i = int(1); i < int(n); i++) {
    int j = f[i - 1];
    while (j > 0 && s[i] != s[j]) j = f[j - 1];
    f[i] = j + (s[i] == s[j]);
  }
  int x = f[n - 1], y = n - x;
  if (x && x % y == 0) return 1;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<vi> a(n);
  while (m--) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    if (x > y) swap(x, y);
    a[x].emplace_back(y - x), a[y].emplace_back(x - y + n);
  }
  string p;
  for (int i = int(0); i < int(n); i++) sort(a[i].begin(), a[i].end());
  for (int i = int(0); i < int(n); i++)
    for (int j : a[i]) p += char(j);
  cout << (hasPeriod(p) ? "Yes\n" : "No\n");
  return 0;
}
