#include <bits/stdc++.h>
using namespace std::chrono;
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
const long long mod = 998244353;
long long n, q, Q, T, m, k, r, x, y, z, g;
set<pair<int, int> > second;
vector<pair<int, int> > plist;
vector<int> divs;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    pair<int, int> x;
    cin >> x.first >> x.second;
    --x.first;
    --x.second;
    if (x.first > x.second) swap(x.first, x.second);
    second.insert(x);
    plist.push_back(x);
  }
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      divs.push_back(i);
      divs.push_back(n / i);
    }
  }
  bool first = 0;
  for (int k : divs) {
    if (k == n) continue;
    bool of = 1;
    for (pair<int, int> x : plist) {
      pair<int, int> o = make_pair((x.first + k) % n, (x.second + k) % n);
      if (o.first > o.second) swap(o.first, o.second);
      if (second.find(o) == second.end()) {
        of = 0;
        break;
      }
    }
    if (of) {
      first = 1;
      break;
    }
  }
  cout << (first ? "Yes" : "No") << endl;
}

### Prompt

Please formulate a cpp solution to 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::chrono;
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
const long long mod = 998244353;
long long n, q, Q, T, m, k, r, x, y, z, g;
set<pair<int, int> > second;
vector<pair<int, int> > plist;
vector<int> divs;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    pair<int, int> x;
    cin >> x.first >> x.second;
    --x.first;
    --x.second;
    if (x.first > x.second) swap(x.first, x.second);
    second.insert(x);
    plist.push_back(x);
  }
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      divs.push_back(i);
      divs.push_back(n / i);
    }
  }
  bool first = 0;
  for (int k : divs) {
    if (k == n) continue;
    bool of = 1;
    for (pair<int, int> x : plist) {
      pair<int, int> o = make_pair((x.first + k) % n, (x.second + k) % n);
      if (o.first > o.second) swap(o.first, o.second);
      if (second.find(o) == second.end()) {
        of = 0;
        break;
      }
    }
    if (of) {
      first = 1;
      break;
    }
  }
  cout << (first ? "Yes" : "No") << endl;
}

```