#include <bits/stdc++.h>
using namespace std;
vector<int> v[100007];
map<vector<int>, int> m;
int n, m1, vis[100007];
bool ch(int x) {
  for (int j = 1; j + x <= n; j++) {
    if (vis[j] != vis[j + x]) return false;
  }
  return true;
}
int cc(int l, int r) {
  if (l > r) {
    swap(l, r);
    int x = ((n - r) + l) % n;
    int y = (r - l);
    if (x > y) return -y;
    return x;
  } else {
    int x = ((n - r) + l) % n;
    int y = (r - l);
    if (x >= y) return y;
    return -x;
  }
}
int main() {
  scanf("%d%d", &n, &m1);
  while (m1--) {
    int l, r;
    scanf("%d%d", &l, &r);
    v[l].push_back(cc(l, r));
    v[r].push_back(cc(r, l));
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    sort(v[i].begin(), v[i].end());
    int x = m[v[i]];
    if (x == 0) {
      x = ++cnt;
      m[v[i]] = x;
    }
    vis[i] = x;
  }
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (i != (n / i) && i > 1) {
        if (ch(n / i)) {
          printf("Yes\n");
          return 0;
        }
      }
      if (ch(i)) {
        printf("Yes\n");
        return 0;
      }
    }
  }
  printf("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;
vector<int> v[100007];
map<vector<int>, int> m;
int n, m1, vis[100007];
bool ch(int x) {
  for (int j = 1; j + x <= n; j++) {
    if (vis[j] != vis[j + x]) return false;
  }
  return true;
}
int cc(int l, int r) {
  if (l > r) {
    swap(l, r);
    int x = ((n - r) + l) % n;
    int y = (r - l);
    if (x > y) return -y;
    return x;
  } else {
    int x = ((n - r) + l) % n;
    int y = (r - l);
    if (x >= y) return y;
    return -x;
  }
}
int main() {
  scanf("%d%d", &n, &m1);
  while (m1--) {
    int l, r;
    scanf("%d%d", &l, &r);
    v[l].push_back(cc(l, r));
    v[r].push_back(cc(r, l));
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    sort(v[i].begin(), v[i].end());
    int x = m[v[i]];
    if (x == 0) {
      x = ++cnt;
      m[v[i]] = x;
    }
    vis[i] = x;
  }
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (i != (n / i) && i > 1) {
        if (ch(n / i)) {
          printf("Yes\n");
          return 0;
        }
      }
      if (ch(i)) {
        printf("Yes\n");
        return 0;
      }
    }
  }
  printf("No\n");
  return 0;
}

```