#include <bits/stdc++.h>
using namespace std;
int st[200005], ed[200005];
set<int> v[100005];
int main() {
  int n, m;
  int x, y;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &x, &y);
    if (x > y) swap(x, y);
    v[x].insert(y);
    st[i] = x;
    ed[i] = y;
  }
  for (int i = 0; i < m; ++i) {
    x = st[i];
    y = ed[i];
    x++;
    y++;
    if (x > n) x -= n;
    if (y > n) y -= n;
    if (x > y) swap(x, y);
    if (!v[x].count(y)) {
      break;
    }
    if (i == m - 1) {
      cout << "Yes" << endl;
      exit(0);
    }
  }
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      for (int j = 0; j < m; ++j) {
        x = st[j];
        y = ed[j];
        x += i;
        y += i;
        if (x > n) x -= n;
        if (y > n) y -= n;
        if (x > y) swap(x, y);
        if (!v[x].count(y)) {
          break;
        }
        if (j == m - 1) {
          cout << "Yes" << endl;
          exit(0);
        }
      }
      for (int j = 0; j < m; ++j) {
        x = st[j];
        y = ed[j];
        x += n / i;
        y += n / i;
        if (x > n) x -= n;
        if (y > n) y -= n;
        if (x > y) swap(x, y);
        if (!v[x].count(y)) {
          break;
        }
        if (j == m - 1) {
          cout << "Yes" << endl;
          exit(0);
        }
      }
    }
  }
  cout << "No" << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp to 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;
int st[200005], ed[200005];
set<int> v[100005];
int main() {
  int n, m;
  int x, y;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &x, &y);
    if (x > y) swap(x, y);
    v[x].insert(y);
    st[i] = x;
    ed[i] = y;
  }
  for (int i = 0; i < m; ++i) {
    x = st[i];
    y = ed[i];
    x++;
    y++;
    if (x > n) x -= n;
    if (y > n) y -= n;
    if (x > y) swap(x, y);
    if (!v[x].count(y)) {
      break;
    }
    if (i == m - 1) {
      cout << "Yes" << endl;
      exit(0);
    }
  }
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      for (int j = 0; j < m; ++j) {
        x = st[j];
        y = ed[j];
        x += i;
        y += i;
        if (x > n) x -= n;
        if (y > n) y -= n;
        if (x > y) swap(x, y);
        if (!v[x].count(y)) {
          break;
        }
        if (j == m - 1) {
          cout << "Yes" << endl;
          exit(0);
        }
      }
      for (int j = 0; j < m; ++j) {
        x = st[j];
        y = ed[j];
        x += n / i;
        y += n / i;
        if (x > n) x -= n;
        if (y > n) y -= n;
        if (x > y) swap(x, y);
        if (!v[x].count(y)) {
          break;
        }
        if (j == m - 1) {
          cout << "Yes" << endl;
          exit(0);
        }
      }
    }
  }
  cout << "No" << endl;
  return 0;
}

```