#include <bits/stdc++.h>
using namespace std;
const int N = 200000 + 5;
int n, m;
pair<int, int> A[N];
set<pair<int, int> > s;
mt19937 rnd(time(0));
bool out;
inline int Getrnd() { return rnd() % n; }
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = (1); i <= (m); ++i) cin >> A[i].first >> A[i].second;
  for (int i = (1); i <= (m); ++i)
    s.insert(A[i]), s.insert(pair<int, int>(A[i].second, A[i].first));
  for (int i = (1); i <= (m); ++i) swap(A[rnd() % m + 1], A[rnd() % m + 1]);
  while ((double)clock() / CLOCKS_PER_SEC <= 2.96) {
    int k = Getrnd();
    if (!k) continue;
    bool Orz = 0;
    for (int i = (1); i <= (m); ++i) {
      if (!s.count(pair<int, int>((A[i].first + k - 1) % n + 1,
                                  (A[i].second + k - 1) % n + 1))) {
        Orz = 1;
        break;
      }
    }
    if (!Orz) {
      puts("Yes");
      out = 1;
      break;
    }
  }
  if (!out) puts("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>
using namespace std;
const int N = 200000 + 5;
int n, m;
pair<int, int> A[N];
set<pair<int, int> > s;
mt19937 rnd(time(0));
bool out;
inline int Getrnd() { return rnd() % n; }
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = (1); i <= (m); ++i) cin >> A[i].first >> A[i].second;
  for (int i = (1); i <= (m); ++i)
    s.insert(A[i]), s.insert(pair<int, int>(A[i].second, A[i].first));
  for (int i = (1); i <= (m); ++i) swap(A[rnd() % m + 1], A[rnd() % m + 1]);
  while ((double)clock() / CLOCKS_PER_SEC <= 2.96) {
    int k = Getrnd();
    if (!k) continue;
    bool Orz = 0;
    for (int i = (1); i <= (m); ++i) {
      if (!s.count(pair<int, int>((A[i].first + k - 1) % n + 1,
                                  (A[i].second + k - 1) % n + 1))) {
        Orz = 1;
        break;
      }
    }
    if (!Orz) {
      puts("Yes");
      out = 1;
      break;
    }
  }
  if (!out) puts("No");
  return 0;
}

```