#include <bits/stdc++.h>
using namespace std;
vector<long long> v[120001], vn;
long long a, b, c[120001], n, m, i, z, k, l, lc, zn;
long long Check(long long d) {
  if (d == v[z].size()) return -1;
  long long i, del = v[z][d] - v[z][0];
  for (i = 0; i < v[z].size(); i++) {
    a = v[z][i];
    b = v[z][(i + d) % v[z].size()];
    if (a < b) swap(a, b);
    if (min(a - b, n - (a - b)) != del) return -1;
  }
  return del;
}
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  while (b != 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    cin >> a >> b;
    if (a < b) swap(a, b);
    v[min(a - b, n - (a - b))].push_back(a);
    v[min(a - b, n - (a - b))].push_back(b);
  }
  for (z = 1; z <= n; z++) {
    if (v[z].size() == 0) continue;
    sort(v[z].begin(), v[z].end());
    vn.clear();
    for (i = 0; i < v[z].size() - 1; i++)
      if (v[z][i] != v[z][i + 1]) vn.push_back(v[z][i]);
    vn.push_back(v[z][v[z].size() - 1]);
    v[z] = vn;
  }
  for (z = 1; z <= n; z++) {
    if (v[z].size() == 0) continue;
    zn = 1000000000000000000LL;
    for (i = 1; i * i <= v[z].size(); i++)
      if (v[z].size() % i == 0) {
        if (Check(i) != -1) zn = min(zn, Check(i));
        if (Check(v[z].size() / i) != -1) zn = min(zn, Check(v[z].size() / i));
      }
    if (zn == 1000000000000000000LL) {
      cout << "No" << '\n';
      return 0;
    }
    k++;
    c[k] = zn;
  }
  l = c[1];
  for (i = 1; i <= k; i++) {
    l = lcm(l, c[i]);
    if (l >= n) break;
  }
  if (l < n)
    cout << "Yes";
  else
    cout << "No";
  cout << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
vector<long long> v[120001], vn;
long long a, b, c[120001], n, m, i, z, k, l, lc, zn;
long long Check(long long d) {
  if (d == v[z].size()) return -1;
  long long i, del = v[z][d] - v[z][0];
  for (i = 0; i < v[z].size(); i++) {
    a = v[z][i];
    b = v[z][(i + d) % v[z].size()];
    if (a < b) swap(a, b);
    if (min(a - b, n - (a - b)) != del) return -1;
  }
  return del;
}
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  while (b != 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    cin >> a >> b;
    if (a < b) swap(a, b);
    v[min(a - b, n - (a - b))].push_back(a);
    v[min(a - b, n - (a - b))].push_back(b);
  }
  for (z = 1; z <= n; z++) {
    if (v[z].size() == 0) continue;
    sort(v[z].begin(), v[z].end());
    vn.clear();
    for (i = 0; i < v[z].size() - 1; i++)
      if (v[z][i] != v[z][i + 1]) vn.push_back(v[z][i]);
    vn.push_back(v[z][v[z].size() - 1]);
    v[z] = vn;
  }
  for (z = 1; z <= n; z++) {
    if (v[z].size() == 0) continue;
    zn = 1000000000000000000LL;
    for (i = 1; i * i <= v[z].size(); i++)
      if (v[z].size() % i == 0) {
        if (Check(i) != -1) zn = min(zn, Check(i));
        if (Check(v[z].size() / i) != -1) zn = min(zn, Check(v[z].size() / i));
      }
    if (zn == 1000000000000000000LL) {
      cout << "No" << '\n';
      return 0;
    }
    k++;
    c[k] = zn;
  }
  l = c[1];
  for (i = 1; i <= k; i++) {
    l = lcm(l, c[i]);
    if (l >= n) break;
  }
  if (l < n)
    cout << "Yes";
  else
    cout << "No";
  cout << '\n';
  return 0;
}

```