#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-6;
set<int> s[100005];
int u[200005], v[200005];
int n, m;
bool check(int k) {
  int a, b;
  int i;
  for (i = 0; i < m; i++) {
    a = (u[i] + k) % n;
    b = (v[i] + k) % n;
    if (a == 0) a = n;
    if (b == 0) b = n;
    if (!s[a].count(b) || !s[b].count(a)) return false;
  }
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  int i;
  for (i = 0; i < m; i++) {
    scanf("%d%d", &u[i], &v[i]);
    s[u[i]].insert(v[i]);
    s[v[i]].insert(u[i]);
  }
  int flag = 0;
  for (i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      flag = check(i);
      if (flag) break;
      if (i != 1 && i * i != n) flag = check(n / i);
      if (flag) break;
    }
  }
  if (flag)
    printf("Yes\n");
  else
    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;
const double EPS = 1e-6;
set<int> s[100005];
int u[200005], v[200005];
int n, m;
bool check(int k) {
  int a, b;
  int i;
  for (i = 0; i < m; i++) {
    a = (u[i] + k) % n;
    b = (v[i] + k) % n;
    if (a == 0) a = n;
    if (b == 0) b = n;
    if (!s[a].count(b) || !s[b].count(a)) return false;
  }
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  int i;
  for (i = 0; i < m; i++) {
    scanf("%d%d", &u[i], &v[i]);
    s[u[i]].insert(v[i]);
    s[v[i]].insert(u[i]);
  }
  int flag = 0;
  for (i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      flag = check(i);
      if (flag) break;
      if (i != 1 && i * i != n) flag = check(n / i);
      if (flag) break;
    }
  }
  if (flag)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}

```