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;
inline void open(const char *s) {}
inline int rd() {
  static int x, f;
  x = 0;
  f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return f > 0 ? x : -x;
}
const int N = 200010;
int n, m;
pair<int, int> a[N];
map<pair<int, int>, bool> mp;
inline bool gao(int d) {
  for (int i = 1; i <= m; i++) {
    int x = (a[i].first - 1 + d) % n + 1;
    int y = (a[i].second - 1 + d) % n + 1;
    if (x > y) swap(x, y);
    if (!mp.count(pair<int, int>(x, y))) return 0;
  }
  return 1;
}
int main() {
  n = rd();
  m = rd();
  for (int i = 1; i <= m; i++) {
    int x = rd(), y = rd();
    if (x > y) swap(x, y);
    mp[a[i] = pair<int, int>(x, y)] = 1;
  }
  for (int i = 1; i <= n - 1; i++)
    if (n % i == 0 && gao(i)) return puts("Yes"), 0;
  puts("No");
  return 0;
}
