#include <bits/stdc++.h>
using namespace std;
vector<int> divisors;
vector<pair<int, int> > edges;
unordered_map<long long, bool> check_shift;
long long f(const long long& a, const long long& b) { return (200000 * a + b); }
bool check(int shift, int n) {
  for (auto e : edges) {
    int a = (e.first + shift) % n;
    int b = (e.second + shift) % n;
    if (a > b) swap(a, b);
    if (check_shift[f(a, b)] == false) return false;
  }
  return true;
}
int main() {
  int n, m, a, b;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    edges.push_back({a - 1, b - 1});
    check_shift[f(a - 1, b - 1)] = true;
  }
  for (int i = 1; i < n; i++)
    if (n % i == 0) divisors.push_back(i);
  for (auto div : divisors) {
    if (check(div, n) == true) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("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;
vector<int> divisors;
vector<pair<int, int> > edges;
unordered_map<long long, bool> check_shift;
long long f(const long long& a, const long long& b) { return (200000 * a + b); }
bool check(int shift, int n) {
  for (auto e : edges) {
    int a = (e.first + shift) % n;
    int b = (e.second + shift) % n;
    if (a > b) swap(a, b);
    if (check_shift[f(a, b)] == false) return false;
  }
  return true;
}
int main() {
  int n, m, a, b;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    edges.push_back({a - 1, b - 1});
    check_shift[f(a - 1, b - 1)] = true;
  }
  for (int i = 1; i < n; i++)
    if (n % i == 0) divisors.push_back(i);
  for (auto div : divisors) {
    if (check(div, n) == true) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

```