#include <bits/stdc++.h>
using PII = std::pair<int, int>;
using LL = long long;
using VI = std::vector<int>;
using CVI = const VI;
using VLL = std::vector<LL>;
using VVI = std::vector<VI>;
using VVLL = std::vector<VLL>;
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<PII> a(m);
  for (auto i = 0; i < (m); ++i) {
    cin >> a[i].first >> a[i].second;
    a.push_back({a[i].second, a[i].first});
  }
  m <<= 1;
  sort((a).begin(), (a).end(), [=](PII& a, PII& b) {
    if (a.first != b.first) return a.first < b.first;
    return (a.second + n - a.first) % n < (b.second + n - b.first) % n;
  });
  VI s;
  for (int j, i = 0; i < m; i = j) {
    for (j = i; j < m && a[j].first == a[i].first; ++j)
      s.push_back((a[j].second + n - a[i].first) % n);
    s.push_back(
        -(j == m ? a[0].first + n - a[i].first : a[j].first - a[i].first));
  }
  m = (int)((s).size());
  VI next(m);
  for (int j, i = 1; i < m; ++i) {
    for (j = next[i - 1]; j > 0 && s[j] != s[i]; j = next[j - 1])
      ;
    next[i] = j + (s[i] == s[j]);
  }
  auto t = s;
  for (auto i = 0; i < (m); ++i) t.push_back(t[i]);
  int i, j;
  for (j = 0, i = 1; i < m * 2 - 1 && j < m;)
    if (t[i] == s[j])
      ++i, ++j;
    else if (j == 0)
      ++i;
    else
      j = next[j - 1];
  cout << (j == m ? "YES" : "NO") << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 PII = std::pair<int, int>;
using LL = long long;
using VI = std::vector<int>;
using CVI = const VI;
using VLL = std::vector<LL>;
using VVI = std::vector<VI>;
using VVLL = std::vector<VLL>;
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<PII> a(m);
  for (auto i = 0; i < (m); ++i) {
    cin >> a[i].first >> a[i].second;
    a.push_back({a[i].second, a[i].first});
  }
  m <<= 1;
  sort((a).begin(), (a).end(), [=](PII& a, PII& b) {
    if (a.first != b.first) return a.first < b.first;
    return (a.second + n - a.first) % n < (b.second + n - b.first) % n;
  });
  VI s;
  for (int j, i = 0; i < m; i = j) {
    for (j = i; j < m && a[j].first == a[i].first; ++j)
      s.push_back((a[j].second + n - a[i].first) % n);
    s.push_back(
        -(j == m ? a[0].first + n - a[i].first : a[j].first - a[i].first));
  }
  m = (int)((s).size());
  VI next(m);
  for (int j, i = 1; i < m; ++i) {
    for (j = next[i - 1]; j > 0 && s[j] != s[i]; j = next[j - 1])
      ;
    next[i] = j + (s[i] == s[j]);
  }
  auto t = s;
  for (auto i = 0; i < (m); ++i) t.push_back(t[i]);
  int i, j;
  for (j = 0, i = 1; i < m * 2 - 1 && j < m;)
    if (t[i] == s[j])
      ++i, ++j;
    else if (j == 0)
      ++i;
    else
      j = next[j - 1];
  cout << (j == m ? "YES" : "NO") << endl;
  return 0;
}

```