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 n, m;
pair<int, int> v[200100], a[200100];
bool check(int k) {
  if (k == n) return false;
  for (int i = 1; i <= m; i++) {
    a[i].first = (v[i].first + k > n ? (v[i].first + k) % n : v[i].first + k),
    a[i].second =
        (v[i].second + k > n ? (v[i].second + k) % n : v[i].second + k);
    if (a[i].first > a[i].second) swap(a[i].first, a[i].second);
  }
  sort(a + 1, a + 1 + m);
  for (int i = 1; i <= m; i++)
    if (a[i].first != v[i].first || a[i].second != v[i].second) return false;
  return true;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d %d", &v[i].first, &v[i].second);
  for (int i = 1; i <= m; i++)
    if (v[i].first > v[i].second) swap(v[i].first, v[i].second);
  sort(v + 1, v + 1 + m);
  for (int k = 1; k * k <= n; k++)
    if (n % k == 0 && (check(k) || check(n / k))) puts("Yes"), exit(0);
  puts("No");
}
