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 N = 1e5 + 10;
int n, m;
vector<int> adj[N];
bool operator==(vector<int> a, vector<int> b) {
  if (a.size() != b.size()) return false;
  for (int i = 0; i < a.size(); i++)
    if (a[i] != b[i]) return false;
  return true;
}
int dist(int x, int y) { return y >= x ? y - x : n - x + y; }
int Z[N];
void CalcZ() {
  Z[1] = n;
  int l = 1, r = 1;
  for (int i = 2; i <= n; i++) {
    if (i > r) {
      l = r = i;
      while (r <= n && adj[r] == adj[r - l + 1]) r++;
      Z[i] = r - l;
      r--;
    } else {
      int k = i - l + 1;
      if (Z[k] < r - i + 1)
        Z[i] = Z[k];
      else {
        l = i;
        while (r <= n && adj[r] == adj[r - l + 1]) r++;
        Z[i] = r - l;
        r--;
      }
    }
  }
}
int main() {
  if (fopen("exam.inp", "r"))
    freopen("exam.inp", "r", stdin), freopen("exam.out", "w", stdout);
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    adj[x].push_back(dist(x, y));
    adj[y].push_back(dist(y, x));
  }
  for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
  CalcZ();
  for (int i = 1; i <= n; i++) Z[i] = (i + Z[i] - 1 == n);
  for (int i = 2; i <= n; i++)
    if (Z[i] && Z[n - i + 2]) {
      puts("Yes");
      return 0;
    }
  puts("No");
}
