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 = 1e6 + 10;
int n, m, x, y;
vector<int> d;
pair<int, int> a[N], b[N];
int main() {
  scanf("%d%d", &n, &m);
  d.push_back(1);
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      d.push_back(i);
      d.push_back(n / i);
    }
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &x, &y);
    x--;
    y--;
    if (x > y) swap(x, y);
    a[i] = make_pair(x, y);
  }
  sort(a + 1, a + m + 1);
  bool f;
  for (auto i : d) {
    for (int j = 1; j <= m; ++j) {
      x = a[j].first + i;
      y = a[j].second + i;
      x %= n;
      y %= n;
      if (x > y) swap(x, y);
      b[j] = make_pair(x, y);
    }
    sort(b + 1, b + m + 1);
    f = 1;
    for (int j = 1; j <= m; ++j) {
      if (a[j] != b[j]) {
        f = 0;
        break;
      }
    }
    if (f) {
      break;
    }
  }
  if (f)
    puts("Yes");
  else
    puts("No");
  return 0;
}
