#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MAX = 1000 * 100 + 47;
vector<int> A[MAX];
int n, m;
int dist(int a, int b) {
  if (b >= a) return b - a;
  return n - a + b;
}
int P[MAX];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = (0); i < (m); i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    if (dist(a, b) > dist(b, a)) {
      swap(a, b);
    }
    A[a].push_back(dist(a, b));
    if (dist(a, b) == dist(b, a)) {
      A[b].push_back(dist(a, b));
    }
  }
  for (int i = (0); i < (n); i++) {
    sort(A[i].begin(), A[i].end());
  }
  for (int i = 1; i < n; ++i) {
    int j = P[i - 1];
    while (j > 0 && A[i] != A[j]) j = P[j - 1];
    if (A[i] == A[j]) ++j;
    P[i] = j;
  }
  int k = n - P[n - 1];
  if (n % k == 0 && k != n) {
    cout << "Yes" << endl;
    return 0;
  }
  cout << "No" << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MAX = 1000 * 100 + 47;
vector<int> A[MAX];
int n, m;
int dist(int a, int b) {
  if (b >= a) return b - a;
  return n - a + b;
}
int P[MAX];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = (0); i < (m); i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    if (dist(a, b) > dist(b, a)) {
      swap(a, b);
    }
    A[a].push_back(dist(a, b));
    if (dist(a, b) == dist(b, a)) {
      A[b].push_back(dist(a, b));
    }
  }
  for (int i = (0); i < (n); i++) {
    sort(A[i].begin(), A[i].end());
  }
  for (int i = 1; i < n; ++i) {
    int j = P[i - 1];
    while (j > 0 && A[i] != A[j]) j = P[j - 1];
    if (A[i] == A[j]) ++j;
    P[i] = j;
  }
  int k = n - P[n - 1];
  if (n % k == 0 && k != n) {
    cout << "Yes" << endl;
    return 0;
  }
  cout << "No" << endl;
  return 0;
}

```