#include <bits/stdc++.h>
using namespace std;
int powmod(int a, int b, int mod) {
  int res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int __set(int b, int i) { return b | (1 << i); }
int __unset(int b, int i) { return b & !(1 << i); }
int __check(int b, int i) { return b & (1 << i); }
const long long inf = 0xFFFFFFFFFFFFFFFL;
long long check(set<pair<long long, long long> >& s,
                vector<pair<long long, long long> >& ls, long long n,
                long long m, long long ch) {
  for (long long i = 0; i < m; i++) {
    pair<long long, long long> l = {(ls[i].first + ch) % n,
                                    (ls[i].second + ch) % n};
    if (s.find({l.first, l.second}) == s.end()) {
      return 0;
    }
  }
  return 1;
}
int32_t main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  long long there[100010] = {};
  memset(there, -1, sizeof(there));
  vector<pair<long long, long long> > ls(m);
  set<pair<long long, long long> > s;
  for (long long i = 0; i < m; i++) {
    long long t1, t2;
    cin >> t1 >> t2;
    s.insert({t1 - 1, t2 - 1});
    s.insert({t2 - 1, t1 - 1});
    ls[i] = {t1 - 1, t2 - 1};
  }
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      long long ch1 = n / i, ch2 = i;
      if (check(s, ls, n, m, ch1) && ch1 != n) {
        cout << "YES"
             << "\n";
        return 0;
      }
      if (check(s, ls, n, m, ch2)) {
        cout << "YES"
             << "\n";
        return 0;
      }
    }
  }
  cout << "NO"
       << "\n";
  return 0;
}

### 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 powmod(int a, int b, int mod) {
  int res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int __set(int b, int i) { return b | (1 << i); }
int __unset(int b, int i) { return b & !(1 << i); }
int __check(int b, int i) { return b & (1 << i); }
const long long inf = 0xFFFFFFFFFFFFFFFL;
long long check(set<pair<long long, long long> >& s,
                vector<pair<long long, long long> >& ls, long long n,
                long long m, long long ch) {
  for (long long i = 0; i < m; i++) {
    pair<long long, long long> l = {(ls[i].first + ch) % n,
                                    (ls[i].second + ch) % n};
    if (s.find({l.first, l.second}) == s.end()) {
      return 0;
    }
  }
  return 1;
}
int32_t main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  long long there[100010] = {};
  memset(there, -1, sizeof(there));
  vector<pair<long long, long long> > ls(m);
  set<pair<long long, long long> > s;
  for (long long i = 0; i < m; i++) {
    long long t1, t2;
    cin >> t1 >> t2;
    s.insert({t1 - 1, t2 - 1});
    s.insert({t2 - 1, t1 - 1});
    ls[i] = {t1 - 1, t2 - 1};
  }
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      long long ch1 = n / i, ch2 = i;
      if (check(s, ls, n, m, ch1) && ch1 != n) {
        cout << "YES"
             << "\n";
        return 0;
      }
      if (check(s, ls, n, m, ch2)) {
        cout << "YES"
             << "\n";
        return 0;
      }
    }
  }
  cout << "NO"
       << "\n";
  return 0;
}

```