#include <bits/stdc++.h>
using namespace std;
int amount, seg;
pair<int, int> segment[200005];
pair<int, int> new_segment[200005];
bool is_good(int roll) {
  if (roll == amount) return false;
  for (int i = 1; i <= seg; i++) {
    new_segment[i].first = (segment[i].first + roll - 1) % amount + 1;
    new_segment[i].second = (segment[i].second + roll - 1) % amount + 1;
    if (new_segment[i].first > new_segment[i].second)
      swap(new_segment[i].first, new_segment[i].second);
  }
  sort(new_segment + 1, new_segment + seg + 1);
  for (int i = 1; i <= seg; i++)
    if (new_segment[i] != segment[i]) return false;
  return true;
}
void solve() {
  for (int i = 1, run_to = (int)sqrt(amount); i <= run_to; i++)
    if (amount % i == 0)
      if (is_good(i) || is_good(amount / i)) {
        printf("Yes");
        return;
      }
  printf("No");
}
int main() {
  scanf("%d %d", &amount, &seg);
  for (int i = 1; i <= seg; i++) {
    scanf("%d %d", &segment[i].first, &segment[i].second);
    if (segment[i].first > segment[i].second)
      swap(segment[i].first, segment[i].second);
  }
  sort(segment + 1, segment + seg + 1);
  solve();
}

### Prompt

In CPP, your task is to solve 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 amount, seg;
pair<int, int> segment[200005];
pair<int, int> new_segment[200005];
bool is_good(int roll) {
  if (roll == amount) return false;
  for (int i = 1; i <= seg; i++) {
    new_segment[i].first = (segment[i].first + roll - 1) % amount + 1;
    new_segment[i].second = (segment[i].second + roll - 1) % amount + 1;
    if (new_segment[i].first > new_segment[i].second)
      swap(new_segment[i].first, new_segment[i].second);
  }
  sort(new_segment + 1, new_segment + seg + 1);
  for (int i = 1; i <= seg; i++)
    if (new_segment[i] != segment[i]) return false;
  return true;
}
void solve() {
  for (int i = 1, run_to = (int)sqrt(amount); i <= run_to; i++)
    if (amount % i == 0)
      if (is_good(i) || is_good(amount / i)) {
        printf("Yes");
        return;
      }
  printf("No");
}
int main() {
  scanf("%d %d", &amount, &seg);
  for (int i = 1; i <= seg; i++) {
    scanf("%d %d", &segment[i].first, &segment[i].second);
    if (segment[i].first > segment[i].second)
      swap(segment[i].first, segment[i].second);
  }
  sort(segment + 1, segment + seg + 1);
  solve();
}

```