#include <bits/stdc++.h>
using namespace std;
long long fastpow(long long a, long long b,
                  long long m = (long long)(1e9 + 7)) {
  long long res = 1;
  a %= m;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
vector<long long> prefix_function(vector<long long> s) {
  long long n = (long long)s.size();
  vector<long long> pi(n);
  for (long long i = 1; i < n; i++) {
    long long j = pi[i - 1];
    while (j > 0 && s[i] != s[j]) j = pi[j - 1];
    if (s[i] == s[j]) j++;
    pi[i] = j;
  }
  return pi;
}
int32_t main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  long long n, m;
  cin >> n >> m;
  set<long long> arr[n + 1];
  for (long long i = 0; i < m; i++) {
    long long u, v;
    cin >> u >> v;
    if (u > v) swap(u, v);
    arr[u].insert(v - u);
    arr[v].insert(u - v + n);
  }
  vector<long long> str;
  for (long long i = 1; i <= n; i++) {
    for (auto x : arr[i]) {
      str.push_back(x);
    }
    str.push_back((long long)9e18);
  }
  long long N = (long long)str.size();
  long long len = prefix_function(str).back();
  if (len > 0 and N % (N - len) == 0) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
  return 0;
}

### 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 fastpow(long long a, long long b,
                  long long m = (long long)(1e9 + 7)) {
  long long res = 1;
  a %= m;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
vector<long long> prefix_function(vector<long long> s) {
  long long n = (long long)s.size();
  vector<long long> pi(n);
  for (long long i = 1; i < n; i++) {
    long long j = pi[i - 1];
    while (j > 0 && s[i] != s[j]) j = pi[j - 1];
    if (s[i] == s[j]) j++;
    pi[i] = j;
  }
  return pi;
}
int32_t main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  long long n, m;
  cin >> n >> m;
  set<long long> arr[n + 1];
  for (long long i = 0; i < m; i++) {
    long long u, v;
    cin >> u >> v;
    if (u > v) swap(u, v);
    arr[u].insert(v - u);
    arr[v].insert(u - v + n);
  }
  vector<long long> str;
  for (long long i = 1; i <= n; i++) {
    for (auto x : arr[i]) {
      str.push_back(x);
    }
    str.push_back((long long)9e18);
  }
  long long N = (long long)str.size();
  long long len = prefix_function(str).back();
  if (len > 0 and N % (N - len) == 0) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
  return 0;
}

```