#include <bits/stdc++.h>
using namespace std;
long long int binomialCoeff(long long int n, long long int k) {
  if (n < k) return 0;
  long long int res = 1;
  if (k > n - k) k = n - k;
  for (int i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long int power(long long int x, long long int y) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
vector<long long int> primes;
vector<bool> prime;
void Sieve(int n) {
  prime = vector<bool>(n + 1, true);
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) primes.push_back(p);
}
const int Nmax = 1e6 + 5;
const int N = 1005;
vector<pair<int, int>> a;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> v;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x %= n;
    y %= n;
    if (x > y) swap(x, y);
    a.push_back({x, y});
  }
  sort(a.begin(), a.end());
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      v.push_back(i);
      if (i != n / i && n / i != n) v.push_back(n / i);
    }
  }
  for (auto x : v) {
    bool yes = true;
    vector<pair<int, int>> w;
    for (auto t : a) {
      int xx = t.first + x;
      int yy = t.second + x;
      xx %= n;
      yy %= n;
      if (xx > yy) swap(xx, yy);
      w.push_back({xx, yy});
    }
    sort(w.begin(), w.end());
    for (int i = 0; i < w.size(); i++) {
      if (w[i] != a[i]) {
        yes = false;
        break;
      }
    }
    if (yes) {
      cout << "yes";
      return 0;
    }
  }
  cout << "no";
}

### 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;
long long int binomialCoeff(long long int n, long long int k) {
  if (n < k) return 0;
  long long int res = 1;
  if (k > n - k) k = n - k;
  for (int i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long int power(long long int x, long long int y) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
vector<long long int> primes;
vector<bool> prime;
void Sieve(int n) {
  prime = vector<bool>(n + 1, true);
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) primes.push_back(p);
}
const int Nmax = 1e6 + 5;
const int N = 1005;
vector<pair<int, int>> a;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> v;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x %= n;
    y %= n;
    if (x > y) swap(x, y);
    a.push_back({x, y});
  }
  sort(a.begin(), a.end());
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      v.push_back(i);
      if (i != n / i && n / i != n) v.push_back(n / i);
    }
  }
  for (auto x : v) {
    bool yes = true;
    vector<pair<int, int>> w;
    for (auto t : a) {
      int xx = t.first + x;
      int yy = t.second + x;
      xx %= n;
      yy %= n;
      if (xx > yy) swap(xx, yy);
      w.push_back({xx, yy});
    }
    sort(w.begin(), w.end());
    for (int i = 0; i < w.size(); i++) {
      if (w[i] != a[i]) {
        yes = false;
        break;
      }
    }
    if (yes) {
      cout << "yes";
      return 0;
    }
  }
  cout << "no";
}

```