#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
int n, m;
vector<pair<int, int> > graph;
vector<int> divi;
int cache[200001];
int b_search(pair<int, int> val) {
  int left = 0, right = m - 1;
  while (left <= right) {
    int mid = (left + right) / 2;
    if (val < graph[mid]) {
      right = mid - 1;
    } else if (val > graph[mid]) {
      left = mid + 1;
    } else
      return mid;
  }
  return -1;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      divi.push_back(i);
      divi.push_back(n / i);
    }
  }
  divi.push_back(1);
  sort(divi.begin(), divi.end());
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    graph.push_back({a - 1, b - 1});
  }
  sort(graph.begin(), graph.end());
  bool ans = false;
  for (int i = 0; i < divi.size(); i++) {
    memset(cache, -1, sizeof(cache));
    bool flag = true;
    for (int j = 0; j < graph.size(); j++) {
      pair<int, int> tmp = graph[j];
      tmp.first = (tmp.first + divi[i]) % n;
      tmp.second = (tmp.second + divi[i]) % n;
      if (tmp.first > tmp.second) swap(tmp.first, tmp.second);
      int res = b_search(tmp);
      if (res == -1) {
        flag = false;
        break;
      }
      if (cache[res] != -1) {
        flag = false;
        break;
      }
      cache[res] = 1;
    }
    if (!flag) continue;
    ans = true;
    break;
  }
  if (ans)
    printf("Yes");
  else
    printf("No");
}

### 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;
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
int n, m;
vector<pair<int, int> > graph;
vector<int> divi;
int cache[200001];
int b_search(pair<int, int> val) {
  int left = 0, right = m - 1;
  while (left <= right) {
    int mid = (left + right) / 2;
    if (val < graph[mid]) {
      right = mid - 1;
    } else if (val > graph[mid]) {
      left = mid + 1;
    } else
      return mid;
  }
  return -1;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      divi.push_back(i);
      divi.push_back(n / i);
    }
  }
  divi.push_back(1);
  sort(divi.begin(), divi.end());
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    graph.push_back({a - 1, b - 1});
  }
  sort(graph.begin(), graph.end());
  bool ans = false;
  for (int i = 0; i < divi.size(); i++) {
    memset(cache, -1, sizeof(cache));
    bool flag = true;
    for (int j = 0; j < graph.size(); j++) {
      pair<int, int> tmp = graph[j];
      tmp.first = (tmp.first + divi[i]) % n;
      tmp.second = (tmp.second + divi[i]) % n;
      if (tmp.first > tmp.second) swap(tmp.first, tmp.second);
      int res = b_search(tmp);
      if (res == -1) {
        flag = false;
        break;
      }
      if (cache[res] != -1) {
        flag = false;
        break;
      }
      cache[res] = 1;
    }
    if (!flag) continue;
    ans = true;
    break;
  }
  if (ans)
    printf("Yes");
  else
    printf("No");
}

```