#include <bits/stdc++.h>
inline int gi() {
  register int x = 0, f = 0;
  register char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
std::set<std::pair<int, int> > st[100010];
std::bitset<200010> bit, I, B, C;
inline bool nmd(int n, int g) {
  for (int i = 1; i < n; ++i)
    for (auto j : st[i]) {
      int a = j.first, b = j.second;
      a = (a + g - 1) % n + 1, b = (b + g - 1) % n + 1;
      if (st[i].find({a, b}) == st[i].end() &&
          st[i].find({b, a}) == st[i].end())
        return 0;
    }
  puts("Yes");
  exit(0);
}
int main() {
  long long n = gi(), m = gi(), a, b, c;
  for (int i = 1; i <= m; ++i) {
    a = gi(), b = gi();
    if (a > b) std::swap(a, b);
    c = std::min(b - a, a + n - b);
    if (b - a != c) std::swap(a, b);
    if (c * 2 == n)
      st[c].insert({b, a}), st[c].insert({a, b});
    else
      st[c].insert({a, b});
  }
  for (int i = 1; i * i <= n; ++i)
    if (n % i == 0) {
      nmd(n, i);
      if (i * i != n && i != 1) nmd(n, n / i);
    }
  puts("No");
  return 0;
}

### Prompt

Your task is to create 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>
inline int gi() {
  register int x = 0, f = 0;
  register char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
std::set<std::pair<int, int> > st[100010];
std::bitset<200010> bit, I, B, C;
inline bool nmd(int n, int g) {
  for (int i = 1; i < n; ++i)
    for (auto j : st[i]) {
      int a = j.first, b = j.second;
      a = (a + g - 1) % n + 1, b = (b + g - 1) % n + 1;
      if (st[i].find({a, b}) == st[i].end() &&
          st[i].find({b, a}) == st[i].end())
        return 0;
    }
  puts("Yes");
  exit(0);
}
int main() {
  long long n = gi(), m = gi(), a, b, c;
  for (int i = 1; i <= m; ++i) {
    a = gi(), b = gi();
    if (a > b) std::swap(a, b);
    c = std::min(b - a, a + n - b);
    if (b - a != c) std::swap(a, b);
    if (c * 2 == n)
      st[c].insert({b, a}), st[c].insert({a, b});
    else
      st[c].insert({a, b});
  }
  for (int i = 1; i * i <= n; ++i)
    if (n % i == 0) {
      nmd(n, i);
      if (i * i != n && i != 1) nmd(n, n / i);
    }
  puts("No");
  return 0;
}

```