#include <bits/stdc++.h>
using namespace std;
map<pair<int, int>, bool> mp;
pair<int, int> ara[400005];
int main() {
  int n, m, x;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &ara[i].first, &ara[i].second);
    ara[i].first %= n, ara[i].second %= n;
    if (ara[i].first > ara[i].second) swap(ara[i].first, ara[i].second);
    ara[i + m] = ara[i], mp[ara[i]] = 1;
  }
  bool f = 0;
  for (int i = 1; i * i <= n && !f; i++)
    if (!(n % i)) {
      x = i;
      bool f2 = 1;
      for (int j = 1; j <= m && f2; j++) {
        int a = (ara[j].first + x) % n;
        int b = (ara[j].second + x) % n;
        if (a > b) swap(a, b);
        f2 = mp[{a, b}];
      }
      f = f2;
      x = (n / i);
      if (x == n || f) continue;
      f2 = 1;
      for (int j = 1; j <= m && f2; j++) {
        int a = (ara[j].first + x) % n;
        int b = (ara[j].second + x) % n;
        if (a > b) swap(a, b);
        f2 = mp[{a, b}];
      }
      f = f2;
    }
  if (f)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
map<pair<int, int>, bool> mp;
pair<int, int> ara[400005];
int main() {
  int n, m, x;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &ara[i].first, &ara[i].second);
    ara[i].first %= n, ara[i].second %= n;
    if (ara[i].first > ara[i].second) swap(ara[i].first, ara[i].second);
    ara[i + m] = ara[i], mp[ara[i]] = 1;
  }
  bool f = 0;
  for (int i = 1; i * i <= n && !f; i++)
    if (!(n % i)) {
      x = i;
      bool f2 = 1;
      for (int j = 1; j <= m && f2; j++) {
        int a = (ara[j].first + x) % n;
        int b = (ara[j].second + x) % n;
        if (a > b) swap(a, b);
        f2 = mp[{a, b}];
      }
      f = f2;
      x = (n / i);
      if (x == n || f) continue;
      f2 = 1;
      for (int j = 1; j <= m && f2; j++) {
        int a = (ara[j].first + x) % n;
        int b = (ara[j].second + x) % n;
        if (a > b) swap(a, b);
        f2 = mp[{a, b}];
      }
      f = f2;
    }
  if (f)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}

```