#include <bits/stdc++.h>
using namespace std;
int n, m;
long long p1 = 100237;
long long p2 = 111119;
long long mod1 = 1000000009;
long long mod2 = 1000000007;
long long pows1[100005];
long long pows2[100005];
vector<int> len[100005];
long long hsh[100005];
long long get_h(vector<int> v) {
  long long res = 0;
  for (int i = 0; i < v.size(); i++) res = (res + v[i] * pows1[i]) % mod1;
  return res;
}
long long sum_hsh;
int main() {
  ios_base::sync_with_stdio(0);
  pows1[0] = 1;
  pows2[0] = 1;
  for (int i = 1; i < 100005; i++) {
    pows1[i] = (pows1[i - 1] * p1) % mod1;
    pows2[i] = (pows2[i - 1] * p2) % mod2;
  }
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b) swap(a, b);
    int lena = b - a;
    int lenb = n - lena;
    len[a].push_back(lena);
    len[b].push_back(lenb);
  }
  for (int i = 1; i <= n; i++) {
    sort(len[i].begin(), len[i].end());
    hsh[i] = get_h(len[i]);
  }
  for (int i = 1; i <= n; i++)
    sum_hsh = (sum_hsh + hsh[i] * pows2[i] % mod2) % mod2;
  long long cur_hsh = sum_hsh;
  for (int i = 0; i < n - 1; i++) {
    cur_hsh = (cur_hsh - hsh[n - i] * pows2[n] % mod2 + mod2) % mod2;
    cur_hsh = (cur_hsh * p2) % mod2;
    cur_hsh = (cur_hsh + hsh[n - i] * pows2[1] % mod2) % mod2;
    if (cur_hsh == sum_hsh) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
}

### 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;
int n, m;
long long p1 = 100237;
long long p2 = 111119;
long long mod1 = 1000000009;
long long mod2 = 1000000007;
long long pows1[100005];
long long pows2[100005];
vector<int> len[100005];
long long hsh[100005];
long long get_h(vector<int> v) {
  long long res = 0;
  for (int i = 0; i < v.size(); i++) res = (res + v[i] * pows1[i]) % mod1;
  return res;
}
long long sum_hsh;
int main() {
  ios_base::sync_with_stdio(0);
  pows1[0] = 1;
  pows2[0] = 1;
  for (int i = 1; i < 100005; i++) {
    pows1[i] = (pows1[i - 1] * p1) % mod1;
    pows2[i] = (pows2[i - 1] * p2) % mod2;
  }
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b) swap(a, b);
    int lena = b - a;
    int lenb = n - lena;
    len[a].push_back(lena);
    len[b].push_back(lenb);
  }
  for (int i = 1; i <= n; i++) {
    sort(len[i].begin(), len[i].end());
    hsh[i] = get_h(len[i]);
  }
  for (int i = 1; i <= n; i++)
    sum_hsh = (sum_hsh + hsh[i] * pows2[i] % mod2) % mod2;
  long long cur_hsh = sum_hsh;
  for (int i = 0; i < n - 1; i++) {
    cur_hsh = (cur_hsh - hsh[n - i] * pows2[n] % mod2 + mod2) % mod2;
    cur_hsh = (cur_hsh * p2) % mod2;
    cur_hsh = (cur_hsh + hsh[n - i] * pows2[1] % mod2) % mod2;
    if (cur_hsh == sum_hsh) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
}

```