#include <bits/stdc++.h>
using namespace std;
int n, m;
set<int> st[101101];
set<int>::iterator it, f;
bool chk(int gap) {
  for (int i = n / 2; i > 0; i--) {
    if (!st[i].size()) continue;
    for (it = st[i].begin(); it != st[i].end(); it++) {
      f = st[i].find(((*it) + gap - 1) % n + 1);
      if (f == st[i].end()) return false;
    }
  }
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    int len = max(x, y) - min(x, y);
    len = min(len, n - len);
    if (y == (x + len - 1) % n + 1) st[len].insert(x);
    if (x == (y + len - 1) % n + 1) st[len].insert(y);
  }
  bool found = false;
  for (int i = 1; i * i <= n && !found; i++) {
    if (n % i) continue;
    found |= chk(i);
    if (!found && i > 1 && i * i < n) found |= chk(n / i);
  }
  puts(found ? "Yes" : "No");
  return 0;
}

### 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<int> st[101101];
set<int>::iterator it, f;
bool chk(int gap) {
  for (int i = n / 2; i > 0; i--) {
    if (!st[i].size()) continue;
    for (it = st[i].begin(); it != st[i].end(); it++) {
      f = st[i].find(((*it) + gap - 1) % n + 1);
      if (f == st[i].end()) return false;
    }
  }
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    int len = max(x, y) - min(x, y);
    len = min(len, n - len);
    if (y == (x + len - 1) % n + 1) st[len].insert(x);
    if (x == (y + len - 1) % n + 1) st[len].insert(y);
  }
  bool found = false;
  for (int i = 1; i * i <= n && !found; i++) {
    if (n % i) continue;
    found |= chk(i);
    if (!found && i > 1 && i * i < n) found |= chk(n / i);
  }
  puts(found ? "Yes" : "No");
  return 0;
}

```