#include <bits/stdc++.h>
using namespace std;
int n, m;
set<pair<int, int> > s;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1, l, r; i <= m; i++) {
    scanf("%d %d", &l, &r);
    if ((r - l + n) % n > n / 2) {
      int p = l;
      l = r;
      r = p;
    }
    s.insert(make_pair(l % n, r % n));
    if (n % 2 == 0 && (r - l + n) % n == n / 2)
      s.insert(make_pair(r % n, l % n));
  }
  for (int p = 1; p * p <= n; p++) {
    if (n % p) continue;
    for (set<pair<int, int> >::iterator it = s.begin(); it != s.end(); it++)
      if (!s.count(make_pair((it->first + p) % n, (it->second + p) % n)))
        goto no1;
    goto yes;
  no1:
    if (p == 1 || p * p == n) continue;
    for (set<pair<int, int> >::iterator it = s.begin(); it != s.end(); it++)
      if (!s.count(
              make_pair((it->first + n / p) % n, (it->second + n / p) % n)))
        goto no2;
    goto yes;
  no2:;
  }
  puts("No");
  return 0;
yes:
  puts("Yes");
}

### 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> > s;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1, l, r; i <= m; i++) {
    scanf("%d %d", &l, &r);
    if ((r - l + n) % n > n / 2) {
      int p = l;
      l = r;
      r = p;
    }
    s.insert(make_pair(l % n, r % n));
    if (n % 2 == 0 && (r - l + n) % n == n / 2)
      s.insert(make_pair(r % n, l % n));
  }
  for (int p = 1; p * p <= n; p++) {
    if (n % p) continue;
    for (set<pair<int, int> >::iterator it = s.begin(); it != s.end(); it++)
      if (!s.count(make_pair((it->first + p) % n, (it->second + p) % n)))
        goto no1;
    goto yes;
  no1:
    if (p == 1 || p * p == n) continue;
    for (set<pair<int, int> >::iterator it = s.begin(); it != s.end(); it++)
      if (!s.count(
              make_pair((it->first + n / p) % n, (it->second + n / p) % n)))
        goto no2;
    goto yes;
  no2:;
  }
  puts("No");
  return 0;
yes:
  puts("Yes");
}

```