#include <bits/stdc++.h>
using namespace std;
int main() {
  long long(n), (m);
  scanf("%lld%lld", &(n), &(m));
  ;
  vector<pair<long long, long long>> A;
  vector<vector<long long>> dists(n + 1);
  for (long long i = 0; i < (m); i++) {
    long long(a), (b);
    scanf("%lld%lld", &(a), &(b));
    ;
    dists[a].push_back((b - a + n) % n);
    dists[b].push_back((a - b + n) % n);
  }
  vector<long long> diffs;
  for (long long i = 1; i < (n + 1); i++) {
    if (dists[i].size() == 0)
      diffs.push_back(-1);
    else {
      sort((dists[i]).begin(), (dists[i]).end());
      for (auto j : dists[i]) diffs.push_back(j);
    }
  }
  vector<long long> s = diffs;
  for (auto i : diffs) s.push_back(i);
  long long sz = (long long)s.size();
  vector<long long> z(sz);
  for (long long i = 1, l = 0, r = 0; i < sz; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < sz && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  for (long long i = 1; i < (diffs.size()); i++) {
    if (z[i] >= diffs.size()) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

### 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() {
  long long(n), (m);
  scanf("%lld%lld", &(n), &(m));
  ;
  vector<pair<long long, long long>> A;
  vector<vector<long long>> dists(n + 1);
  for (long long i = 0; i < (m); i++) {
    long long(a), (b);
    scanf("%lld%lld", &(a), &(b));
    ;
    dists[a].push_back((b - a + n) % n);
    dists[b].push_back((a - b + n) % n);
  }
  vector<long long> diffs;
  for (long long i = 1; i < (n + 1); i++) {
    if (dists[i].size() == 0)
      diffs.push_back(-1);
    else {
      sort((dists[i]).begin(), (dists[i]).end());
      for (auto j : dists[i]) diffs.push_back(j);
    }
  }
  vector<long long> s = diffs;
  for (auto i : diffs) s.push_back(i);
  long long sz = (long long)s.size();
  vector<long long> z(sz);
  for (long long i = 1, l = 0, r = 0; i < sz; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < sz && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  for (long long i = 1; i < (diffs.size()); i++) {
    if (z[i] >= diffs.size()) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

```