#include <bits/stdc++.h>
using namespace std;
const long long llINF = 1e18, llMOD = 1e9 + 7;
const int INF = 1e9, MOD = 1e9 + 7, P = 29, N = 2e5 + 2, K = 200, L = 20,
          A = 256;
const long double EPS = 1e-9, ldINF = 1e18, PI = acos(-1);
string sep = " ";
template <typename T, typename P>
istream &operator>>(istream &in, pair<T, P> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T>
void in_seg(istream &in, T *dst_begin, T *dst_end) {
  while (dst_begin != dst_end) {
    in >> (*dst_begin);
    ++dst_begin;
  }
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (int i = 0; i < v.size(); ++i) in >> v[i];
  return in;
}
template <typename T, typename P>
ostream &operator<<(ostream &out, pair<T, P> p) {
  out << p.first << ' ' << p.second;
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
  for (int i = 0; i < v.size(); ++i) {
    out << v[i];
    if (i != v.size() - 1) out << sep;
  }
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, vector<vector<T>> &v) {
  for (int i = 0; i < v.size(); ++i) {
    out << v[i];
    if (i != v.size() - 1) out << '\n';
  }
  return out;
}
void set_precision(ostream &out, int val) {
  out.setf(out.fixed, out.floatfield);
  out.precision(val);
}
unordered_set<long long> st;
int main() {
  st.rehash(1e6);
  long long n, m;
  cin >> n >> m;
  vector<pair<int, int>> seg;
  for (int i = 0, l, r; i < m; i++) {
    cin >> l >> r;
    --l, --r;
    st.insert(n * l + r);
    st.insert(n * r + l);
    seg.push_back({l, r});
  }
  for (long long k = 1; k < n; k++) {
    if (n % k) continue;
    bool flag = true;
    for (auto [l, r] : seg)
      flag &= (st.find(((l + k) % n) * n + (r + k) % n) != st.end());
    if (flag) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No";
}

### Prompt

In CPP, your task is to solve 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;
const long long llINF = 1e18, llMOD = 1e9 + 7;
const int INF = 1e9, MOD = 1e9 + 7, P = 29, N = 2e5 + 2, K = 200, L = 20,
          A = 256;
const long double EPS = 1e-9, ldINF = 1e18, PI = acos(-1);
string sep = " ";
template <typename T, typename P>
istream &operator>>(istream &in, pair<T, P> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T>
void in_seg(istream &in, T *dst_begin, T *dst_end) {
  while (dst_begin != dst_end) {
    in >> (*dst_begin);
    ++dst_begin;
  }
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (int i = 0; i < v.size(); ++i) in >> v[i];
  return in;
}
template <typename T, typename P>
ostream &operator<<(ostream &out, pair<T, P> p) {
  out << p.first << ' ' << p.second;
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
  for (int i = 0; i < v.size(); ++i) {
    out << v[i];
    if (i != v.size() - 1) out << sep;
  }
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, vector<vector<T>> &v) {
  for (int i = 0; i < v.size(); ++i) {
    out << v[i];
    if (i != v.size() - 1) out << '\n';
  }
  return out;
}
void set_precision(ostream &out, int val) {
  out.setf(out.fixed, out.floatfield);
  out.precision(val);
}
unordered_set<long long> st;
int main() {
  st.rehash(1e6);
  long long n, m;
  cin >> n >> m;
  vector<pair<int, int>> seg;
  for (int i = 0, l, r; i < m; i++) {
    cin >> l >> r;
    --l, --r;
    st.insert(n * l + r);
    st.insert(n * r + l);
    seg.push_back({l, r});
  }
  for (long long k = 1; k < n; k++) {
    if (n % k) continue;
    bool flag = true;
    for (auto [l, r] : seg)
      flag &= (st.find(((l + k) % n) * n + (r + k) % n) != st.end());
    if (flag) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No";
}

```