#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  set<int> li[n];
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    int t = min(a, b);
    b = max(a, b);
    a = t;
    li[a].insert(b - a);
    li[b].insert(n + a - b);
  }
  int s = (n + 2 * m) * 2 - 1;
  int nee[s];
  int cc = 0;
  for (int i = 0; i < n; i++) {
    nee[cc] = 0;
    cc++;
    for (set<int>::iterator it = li[i].begin(); it != li[i].end(); it++, cc++) {
      nee[cc] = *it;
    }
  }
  for (int i = n + 2 * m; i < s; i++) nee[i] = nee[i - (n + 2 * m)];
  int kmp[s];
  kmp[0] = -1;
  bool fo = 0;
  for (int i = 1; i < s; i++) {
    int p1 = i - 1, p2 = kmp[p1];
    while (nee[i] != nee[p2 + 1] && p2 != -1) {
      p1 = p2;
      p2 = kmp[p2];
    }
    if (nee[i] == nee[p2 + 1])
      kmp[i] = kmp[p1] + 1;
    else
      kmp[i] = -1;
    if (kmp[i] == n + 2 * m - 1) {
      fo = 1;
      break;
    }
  }
  if (fo)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 main() {
  int n, m;
  cin >> n >> m;
  set<int> li[n];
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    int t = min(a, b);
    b = max(a, b);
    a = t;
    li[a].insert(b - a);
    li[b].insert(n + a - b);
  }
  int s = (n + 2 * m) * 2 - 1;
  int nee[s];
  int cc = 0;
  for (int i = 0; i < n; i++) {
    nee[cc] = 0;
    cc++;
    for (set<int>::iterator it = li[i].begin(); it != li[i].end(); it++, cc++) {
      nee[cc] = *it;
    }
  }
  for (int i = n + 2 * m; i < s; i++) nee[i] = nee[i - (n + 2 * m)];
  int kmp[s];
  kmp[0] = -1;
  bool fo = 0;
  for (int i = 1; i < s; i++) {
    int p1 = i - 1, p2 = kmp[p1];
    while (nee[i] != nee[p2 + 1] && p2 != -1) {
      p1 = p2;
      p2 = kmp[p2];
    }
    if (nee[i] == nee[p2 + 1])
      kmp[i] = kmp[p1] + 1;
    else
      kmp[i] = -1;
    if (kmp[i] == n + 2 * m - 1) {
      fo = 1;
      break;
    }
  }
  if (fo)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

```