#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y, long long md = 1000000007) {
  long long res = 1;
  x %= md;
  while (y > 0) {
    if (y & 1) res = (res * x) % md;
    x = (x * x) % md;
    y = y >> 1;
  }
  return res % md;
}
signed main() {
  long long n, m;
  cin >> n >> m;
  long long root = sqrt(n) + 1;
  std::vector<long long> factors;
  std::vector<pair<int, int> > v(m);
  for (int i = 1; i < root; i++) {
    if (n % i) continue;
    long long temp = n / i;
    factors.push_back(i);
    if (i != temp && temp != n) factors.push_back(temp);
  }
  std::vector<long long> check(m, 0);
  std::map<pair<long long, long long>, long long> map;
  for (int i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    if (b < a) swap(a, b);
    v[i] = make_pair(a, b);
    map[v[i]] = i;
  }
  for (int i = 0; i < factors.size(); i++) {
    for (int j = 0; j < m; j++) check[i] = 0;
    long long x = factors[i];
    long long first = 1;
    for (int j = 0; j < m; j++) {
      long long a = min((v[j].first + x) % n, (v[j].second + x) % n);
      long long b = max((v[j].first + x) % n, (v[j].second + x) % n);
      std::map<pair<long long, long long>, long long>::iterator it =
          map.find(make_pair(a, b));
      if (it == map.end()) {
        first = 0;
        break;
      }
    }
    if (first) {
      cout << "Yes";
      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;
long long power(long long x, long long y, long long md = 1000000007) {
  long long res = 1;
  x %= md;
  while (y > 0) {
    if (y & 1) res = (res * x) % md;
    x = (x * x) % md;
    y = y >> 1;
  }
  return res % md;
}
signed main() {
  long long n, m;
  cin >> n >> m;
  long long root = sqrt(n) + 1;
  std::vector<long long> factors;
  std::vector<pair<int, int> > v(m);
  for (int i = 1; i < root; i++) {
    if (n % i) continue;
    long long temp = n / i;
    factors.push_back(i);
    if (i != temp && temp != n) factors.push_back(temp);
  }
  std::vector<long long> check(m, 0);
  std::map<pair<long long, long long>, long long> map;
  for (int i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    if (b < a) swap(a, b);
    v[i] = make_pair(a, b);
    map[v[i]] = i;
  }
  for (int i = 0; i < factors.size(); i++) {
    for (int j = 0; j < m; j++) check[i] = 0;
    long long x = factors[i];
    long long first = 1;
    for (int j = 0; j < m; j++) {
      long long a = min((v[j].first + x) % n, (v[j].second + x) % n);
      long long b = max((v[j].first + x) % n, (v[j].second + x) % n);
      std::map<pair<long long, long long>, long long>::iterator it =
          map.find(make_pair(a, b));
      if (it == map.end()) {
        first = 0;
        break;
      }
    }
    if (first) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
}

```