#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
int n, m;
int p[maxn], l[maxn], r[maxn];
pair<int, int> b[maxn], e[maxn];
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> l[i] >> r[i];
    int u = l[i], v = r[i];
    if (u > v) swap(u, v);
    b[i] = pair<int, int>(u, v);
  }
  sort(b + 1, b + m + 1);
  int sl = 0;
  for (int i = 1; i <= sqrt(n); ++i)
    if (n % i == 0) {
      p[++sl] = i;
      if (i != n / i && n / i != n) p[++sl] = n / i;
    }
  for (int k = 1; k <= sl; ++k) {
    for (int i = 1; i <= m; ++i) {
      int u = (l[i] + p[k] - 1) % n + 1;
      int v = (r[i] + p[k] - 1) % n + 1;
      if (u > v) swap(u, v);
      e[i] = pair<int, int>(u, v);
    }
    sort(e + 1, e + m + 1);
    for (int i = 1; i <= m; ++i)
      if (e[i] != b[i]) goto l1;
    puts("Yes");
    return;
  l1:;
  }
  puts("No");
}
int main() { solve(); }

### Prompt

Construct a CPP code solution to the problem outlined:

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 maxn = 1e6 + 5;
int n, m;
int p[maxn], l[maxn], r[maxn];
pair<int, int> b[maxn], e[maxn];
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> l[i] >> r[i];
    int u = l[i], v = r[i];
    if (u > v) swap(u, v);
    b[i] = pair<int, int>(u, v);
  }
  sort(b + 1, b + m + 1);
  int sl = 0;
  for (int i = 1; i <= sqrt(n); ++i)
    if (n % i == 0) {
      p[++sl] = i;
      if (i != n / i && n / i != n) p[++sl] = n / i;
    }
  for (int k = 1; k <= sl; ++k) {
    for (int i = 1; i <= m; ++i) {
      int u = (l[i] + p[k] - 1) % n + 1;
      int v = (r[i] + p[k] - 1) % n + 1;
      if (u > v) swap(u, v);
      e[i] = pair<int, int>(u, v);
    }
    sort(e + 1, e + m + 1);
    for (int i = 1; i <= m; ++i)
      if (e[i] != b[i]) goto l1;
    puts("Yes");
    return;
  l1:;
  }
  puts("No");
}
int main() { solve(); }

```