#include <bits/stdc++.h>
using namespace std;
const int N = 100010, Mod = 1e9 + 7;
int n, m, x, y;
set<int> adj[N];
bool check(int x) {
  for (int i = 0; i < n; i++) {
    if (adj[(i + x) % n].size() != adj[i].size()) return false;
    for (auto v : adj[i]) {
      if (adj[(i + x) % n].count((v + x) % n) == 0) return false;
    }
  }
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    --x, --y;
    adj[x].insert(y);
    adj[y].insert(x);
  }
  for (int i = 1; i * i <= n; i++) {
    if (i != 1) {
      if (check(n / i)) {
        printf("Yes\n");
        return 0;
      }
    }
    if (check(i)) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
const int N = 100010, Mod = 1e9 + 7;
int n, m, x, y;
set<int> adj[N];
bool check(int x) {
  for (int i = 0; i < n; i++) {
    if (adj[(i + x) % n].size() != adj[i].size()) return false;
    for (auto v : adj[i]) {
      if (adj[(i + x) % n].count((v + x) % n) == 0) return false;
    }
  }
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    --x, --y;
    adj[x].insert(y);
    adj[y].insert(x);
  }
  for (int i = 1; i * i <= n; i++) {
    if (i != 1) {
      if (check(n / i)) {
        printf("Yes\n");
        return 0;
      }
    }
    if (check(i)) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

```