#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-11;
template <class T>
inline void ckmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void ckmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
using VI = vector<int>;
int main() {
  std::ios::sync_with_stdio(false);
  int n, m;
  while (cin >> n >> m) {
    vector<VI> g(n);
    for (int i = 0; i < (m); ++i) {
      int a, b;
      cin >> a >> b;
      --a;
      --b;
      g[a].push_back((b - a + n) % n);
      g[b].push_back((a - b + n) % n);
    }
    for (int i = 0; i < (n); ++i) sort(g[i].begin(), g[i].end());
    int ok = 0;
    for (int d = 1; d < n; d++)
      if (n % d == 0) {
        int flag = 1;
        for (int i = 0; i < (n); ++i)
          if (g[i] != g[(i + d) % n]) {
            flag = 0;
            break;
          }
        if (flag) {
          ok = 1;
          break;
        }
      }
    if (ok)
      printf("Yes\n");
    else
      printf("No\n");
    break;
  }
  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;
const double pi = acos(-1.0);
const double eps = 1e-11;
template <class T>
inline void ckmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void ckmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
using VI = vector<int>;
int main() {
  std::ios::sync_with_stdio(false);
  int n, m;
  while (cin >> n >> m) {
    vector<VI> g(n);
    for (int i = 0; i < (m); ++i) {
      int a, b;
      cin >> a >> b;
      --a;
      --b;
      g[a].push_back((b - a + n) % n);
      g[b].push_back((a - b + n) % n);
    }
    for (int i = 0; i < (n); ++i) sort(g[i].begin(), g[i].end());
    int ok = 0;
    for (int d = 1; d < n; d++)
      if (n % d == 0) {
        int flag = 1;
        for (int i = 0; i < (n); ++i)
          if (g[i] != g[(i + d) % n]) {
            flag = 0;
            break;
          }
        if (flag) {
          ok = 1;
          break;
        }
      }
    if (ok)
      printf("Yes\n");
    else
      printf("No\n");
    break;
  }
  return 0;
}

```