#include <bits/stdc++.h>
using namespace std;
const int N = 500005;
int n, ans, m, p[N];
set<pair<int, int> > s;
bool check(int x) {
  for (auto p : s) {
    int u = (p.first + x) % n, v = (p.second + x) % n;
    if (u > v) swap(u, v);
    if (s.find(make_pair(u, v)) == s.end()) return 0;
  }
  return 1;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - '0';
  return x * f;
}
int main() {
  for (int i = 2; i < N; i++)
    if (!p[i])
      for (int j = 2; j * i < N; j++) p[i * j] = 1;
  n = read();
  m = read();
  while (m--) {
    int u = read() - 1, v = read() - 1;
    if (u > v) swap(u, v);
    s.insert(make_pair(u, v));
  }
  for (int i = 1; i * i <= n; i++)
    if (!(n % i)) {
      if (check(i)) {
        puts("Yes");
        return 0;
      } else if (i > 1 && check(n / i)) {
        puts("Yes");
        return 0;
      }
    }
  puts("No");
  return 0;
}

### Prompt

Please formulate 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 = 500005;
int n, ans, m, p[N];
set<pair<int, int> > s;
bool check(int x) {
  for (auto p : s) {
    int u = (p.first + x) % n, v = (p.second + x) % n;
    if (u > v) swap(u, v);
    if (s.find(make_pair(u, v)) == s.end()) return 0;
  }
  return 1;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - '0';
  return x * f;
}
int main() {
  for (int i = 2; i < N; i++)
    if (!p[i])
      for (int j = 2; j * i < N; j++) p[i * j] = 1;
  n = read();
  m = read();
  while (m--) {
    int u = read() - 1, v = read() - 1;
    if (u > v) swap(u, v);
    s.insert(make_pair(u, v));
  }
  for (int i = 1; i * i <= n; i++)
    if (!(n % i)) {
      if (check(i)) {
        puts("Yes");
        return 0;
      } else if (i > 1 && check(n / i)) {
        puts("Yes");
        return 0;
      }
    }
  puts("No");
  return 0;
}

```