#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000000 + 10;
const int oo = 1 << 30;
vector<int> graph[maxn];
int N, M, b[maxn];
void pre(vector<int> P) {
  int i = 0, j = -1;
  b[0] = -1;
  while (i < P.size()) {
    while (j >= 0 && P[i] != P[j]) j = b[j];
    i++;
    j++;
    b[i] = j;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  for (int i = 0; i < M; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  vector<int> txt;
  for (int i = 1; i < N + 1; i++) {
    txt.push_back(-1);
    vector<int> dist;
    for (auto &x : graph[i]) {
      int d = (x < i) ? (N - i + x) : x - i;
      dist.push_back(d);
    }
    sort((dist).begin(), (dist).end());
    for (auto &x : dist) txt.push_back(x);
  }
  pre(txt);
  int act = txt.size();
  int mini = oo;
  int d = txt.size() - b[act];
  if (d && txt.size() % d == 0)
    mini = d;
  else
    mini = txt.size();
  int p = txt.size() / mini;
  if (p > 1)
    cout << "Yes\n";
  else
    cout << "No\n";
  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 int maxn = 1000000 + 10;
const int oo = 1 << 30;
vector<int> graph[maxn];
int N, M, b[maxn];
void pre(vector<int> P) {
  int i = 0, j = -1;
  b[0] = -1;
  while (i < P.size()) {
    while (j >= 0 && P[i] != P[j]) j = b[j];
    i++;
    j++;
    b[i] = j;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  for (int i = 0; i < M; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  vector<int> txt;
  for (int i = 1; i < N + 1; i++) {
    txt.push_back(-1);
    vector<int> dist;
    for (auto &x : graph[i]) {
      int d = (x < i) ? (N - i + x) : x - i;
      dist.push_back(d);
    }
    sort((dist).begin(), (dist).end());
    for (auto &x : dist) txt.push_back(x);
  }
  pre(txt);
  int act = txt.size();
  int mini = oo;
  int d = txt.size() - b[act];
  if (d && txt.size() % d == 0)
    mini = d;
  else
    mini = txt.size();
  int p = txt.size() / mini;
  if (p > 1)
    cout << "Yes\n";
  else
    cout << "No\n";
  return 0;
}

```