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 a[200010], b[200010];
vector<pair<int, int> > v;
vector<pair<int, int> > aux;
int n, m;
int main() {
  scanf("%d %d", &n, &m);
  int i, j;
  for (i = 1; i <= m; i++) {
    scanf("%d %d", &a[i], &b[i]);
    if (a[i] > b[i]) {
      swap(a[i], b[i]);
    }
    a[i]--;
    b[i]--;
    v.push_back(make_pair(a[i], b[i]));
  }
  sort(v.begin(), v.end());
  int pode = 0;
  int k;
  for (k = 1; k < n; k++) {
    if (n % k == 0) {
      j = 0;
      if (j < v.size()) {
        aux.resize(0);
        for (i = j; i < v.size(); i++) {
          aux.push_back(make_pair((v[i].first + k) % n, (v[i].second + k) % n));
        }
        for (i = 0; i < j; i++) {
          aux.push_back(make_pair((v[i].first + k) % n, (v[i].second + k) % n));
        }
        for (i = 0; i < aux.size(); i++) {
          if (aux[i].first > aux[i].second) {
            swap(aux[i].first, aux[i].second);
          }
        }
        sort(aux.begin(), aux.end());
        if (v == aux) {
          pode = 1;
        }
      }
    }
  }
  if (pode == 1) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }
  return 0;
}
