#include <bits/stdc++.h>
using namespace std;
int n, m;
set<pair<int, int> > G;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (u > v) swap(u, v);
    G.insert(make_pair(u, v));
  }
  bool fa = 0;
  set<pair<int, int> >::iterator it;
  for (int i = 1; i < n; i++) {
    if (n % i) continue;
    fa = 1;
    for (it = G.begin(); it != G.end(); it++) {
      int u = (it->first + i) % n;
      int v = (it->second + i) % n;
      if (u == 0) u = n;
      if (v == 0) v = n;
      if (u > v) swap(u, v);
      if (!G.count(make_pair(u, v))) {
        fa = 0;
        break;
      }
    }
    if (fa) break;
  }
  if (fa)
    puts("YES");
  else
    puts("NO");
}

### Prompt

Generate a Cpp solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m;
set<pair<int, int> > G;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (u > v) swap(u, v);
    G.insert(make_pair(u, v));
  }
  bool fa = 0;
  set<pair<int, int> >::iterator it;
  for (int i = 1; i < n; i++) {
    if (n % i) continue;
    fa = 1;
    for (it = G.begin(); it != G.end(); it++) {
      int u = (it->first + i) % n;
      int v = (it->second + i) % n;
      if (u == 0) u = n;
      if (v == 0) v = n;
      if (u > v) swap(u, v);
      if (!G.count(make_pair(u, v))) {
        fa = 0;
        break;
      }
    }
    if (fa) break;
  }
  if (fa)
    puts("YES");
  else
    puts("NO");
}

```