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;
int n, m;
vector<pii> a;
unordered_set<ll> s;
bool f(int r) {
  for (int i = int(0); i < int(m); i++) {
    ll x = a[i].first + r, y = a[i].second + r;
    if (x >= n) x -= n;
    if (y >= n) y -= n;
    if (x > y) swap(x, y);
    if (!s.count(x * n + y)) return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  a.resize(m);
  s.reserve(1 << 18);
  s.max_load_factor(0.25);
  for (int i = int(0); i < int(m); i++) {
    int x, y;
    cin >> x >> y;
    if (x == n) x = 0;
    if (y == n) y = 0;
    if (x > y) swap(x, y);
    a[i] = {x, y};
    s.emplace(ll(x) * n + y);
  }
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0 && (f(i) || (i > 1 && f(n / i)))) return cout << "Yes\n", 0;
  }
  return cout << "No\n", 0;
}
