#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 69, MOD = 1e9 + 7;
int n, m, cnt, v[MAXN], u[MAXN], divs[MAXN >> 1];
unordered_map<int, bool> mp;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> v[i] >> u[i], v[i]--, u[i]--;
    if (v[i] < u[i]) swap(v[i], u[i]);
    mp[MAXN * v[i] + u[i]] = true;
  }
  divs[cnt++] = 1;
  for (int it = 2; it <= sqrt(n); it++)
    if (n % it == 0) divs[cnt++] = it, divs[cnt++] = n / it;
  for (int it = 0; it < cnt; it++) {
    bool exist = true;
    for (int i = 0; i < m; i++) {
      int V = (v[i] + divs[it]) % n, U = (u[i] + divs[it]) % n;
      if (V < U) swap(V, U);
      exist &= mp[MAXN * V + U];
      if (m * it + sqrt(n) > 6000000) return cout << "No\n", 0;
    }
    if (exist) return cout << "Yes\n", 0;
  }
  cout << "No\n";
}

### Prompt

Create a solution in cpp for 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 MAXN = 2e5 + 69, MOD = 1e9 + 7;
int n, m, cnt, v[MAXN], u[MAXN], divs[MAXN >> 1];
unordered_map<int, bool> mp;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> v[i] >> u[i], v[i]--, u[i]--;
    if (v[i] < u[i]) swap(v[i], u[i]);
    mp[MAXN * v[i] + u[i]] = true;
  }
  divs[cnt++] = 1;
  for (int it = 2; it <= sqrt(n); it++)
    if (n % it == 0) divs[cnt++] = it, divs[cnt++] = n / it;
  for (int it = 0; it < cnt; it++) {
    bool exist = true;
    for (int i = 0; i < m; i++) {
      int V = (v[i] + divs[it]) % n, U = (u[i] + divs[it]) % n;
      if (V < U) swap(V, U);
      exist &= mp[MAXN * V + U];
      if (m * it + sqrt(n) > 6000000) return cout << "No\n", 0;
    }
    if (exist) return cout << "Yes\n", 0;
  }
  cout << "No\n";
}

```