#include <bits/stdc++.h>
const int N = 2e5 + 10;
int mod, n, a, b, good;
std::pair<int, int> set[N];
std::mt19937 rng(std::chrono::steady_clock::now().time_since_epoch().count());
std::pair<int, int> make_pair(int i, int j) {
  return {std::min(i, j), std::max(i, j)};
}
int main() {
  scanf("%d%d", &mod, &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a, &b);
    set[i] = make_pair(a - 1, b - 1);
  }
  shuffle(set, set + n, rng);
  std::sort(set, set + n);
  for (int k = 1; k < mod; k++) {
    if (mod % k) continue;
    good = true;
    for (auto p = set; p != set + n; p++)
      if (!std::binary_search(
              set, set + n,
              make_pair((p->first + k) % mod, (p->second + k) % mod))) {
        good = false;
        break;
      }
    if (good) break;
  }
  printf("%s\n", good ? "yes" : "no");
  return 0;
}

### 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>
const int N = 2e5 + 10;
int mod, n, a, b, good;
std::pair<int, int> set[N];
std::mt19937 rng(std::chrono::steady_clock::now().time_since_epoch().count());
std::pair<int, int> make_pair(int i, int j) {
  return {std::min(i, j), std::max(i, j)};
}
int main() {
  scanf("%d%d", &mod, &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a, &b);
    set[i] = make_pair(a - 1, b - 1);
  }
  shuffle(set, set + n, rng);
  std::sort(set, set + n);
  for (int k = 1; k < mod; k++) {
    if (mod % k) continue;
    good = true;
    for (auto p = set; p != set + n; p++)
      if (!std::binary_search(
              set, set + n,
              make_pair((p->first + k) % mod, (p->second + k) % mod))) {
        good = false;
        break;
      }
    if (good) break;
  }
  printf("%s\n", good ? "yes" : "no");
  return 0;
}

```