#include <bits/stdc++.h>
using namespace std;
double PI = 4.00 * (atan(1));
pair<long long int, long long int> lines[200010];
long long int number, segment;
map<pair<long long int, long long int>, long long int> exist;
bool jumpcheck(long long int jump) {
  bool check = true;
  pair<long long int, long long int> temp;
  for (long long int i = 0; i < segment; i++) {
    temp.first = (lines[i].first - 1 + jump) % number;
    temp.first++;
    temp.second = (lines[i].second - 1 + jump) % number;
    temp.second++;
    if (exist[temp] == 0) {
      check = false;
      break;
    }
  }
  return check;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> number >> segment;
  pair<long long int, long long int> temp;
  for (long long int i = 0; i < segment; i++) {
    cin >> lines[i].first >> lines[i].second;
    temp = {lines[i].second, lines[i].first};
    exist[lines[i]] = exist[temp] = 1;
  }
  set<long long int> divisor;
  for (long long int i = 1; i * i <= number; i++) {
    if (number % i == 0) {
      divisor.insert(i);
      divisor.insert(number / i);
    }
  }
  vector<long long int> final(divisor.begin(), divisor.end());
  long long int size = final.size();
  for (long long int i = 0; i < size - 1; i++) {
    if (jumpcheck(final[i])) {
      cout << "Yes";
      exit(0);
    }
  }
  cout << "No";
  return 0;
}

### Prompt

Generate 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;
double PI = 4.00 * (atan(1));
pair<long long int, long long int> lines[200010];
long long int number, segment;
map<pair<long long int, long long int>, long long int> exist;
bool jumpcheck(long long int jump) {
  bool check = true;
  pair<long long int, long long int> temp;
  for (long long int i = 0; i < segment; i++) {
    temp.first = (lines[i].first - 1 + jump) % number;
    temp.first++;
    temp.second = (lines[i].second - 1 + jump) % number;
    temp.second++;
    if (exist[temp] == 0) {
      check = false;
      break;
    }
  }
  return check;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> number >> segment;
  pair<long long int, long long int> temp;
  for (long long int i = 0; i < segment; i++) {
    cin >> lines[i].first >> lines[i].second;
    temp = {lines[i].second, lines[i].first};
    exist[lines[i]] = exist[temp] = 1;
  }
  set<long long int> divisor;
  for (long long int i = 1; i * i <= number; i++) {
    if (number % i == 0) {
      divisor.insert(i);
      divisor.insert(number / i);
    }
  }
  vector<long long int> final(divisor.begin(), divisor.end());
  long long int size = final.size();
  for (long long int i = 0; i < size - 1; i++) {
    if (jumpcheck(final[i])) {
      cout << "Yes";
      exit(0);
    }
  }
  cout << "No";
  return 0;
}

```