#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr double kPI = 3.14159265358979323846264338327950288;
vector<ll> GetAllDivisorsUnsorted(ll a) {
  vector<ll> divs;
  for (ll i = 1; i * i <= a; ++i) {
    if (a % i == 0) {
      divs.push_back(i);
      if (i != a / i) divs.push_back(a / i);
    }
  }
  return divs;
}
int n, m;
vector<int> a, b;
set<pair<int, int>> segs;
void Load() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    segs.insert({a, b});
    segs.insert({b, a});
  }
}
bool RotateWorks(ll div) {
  for (const auto& seg : segs) {
    if (segs.count({(seg.first + div) % n, (seg.second + div) % n}) == 0)
      return false;
  }
  return true;
}
void Solve() {
  vector<ll> divs = GetAllDivisorsUnsorted(n);
  for (ll div : divs) {
    if (div == n) continue;
    if (RotateWorks(div)) {
      puts("Yes");
      return;
    }
  }
  puts("No");
}
int main() {
  Load();
  Solve();
}

### Prompt

Please provide a Cpp coded solution 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;
using ll = long long;
constexpr double kPI = 3.14159265358979323846264338327950288;
vector<ll> GetAllDivisorsUnsorted(ll a) {
  vector<ll> divs;
  for (ll i = 1; i * i <= a; ++i) {
    if (a % i == 0) {
      divs.push_back(i);
      if (i != a / i) divs.push_back(a / i);
    }
  }
  return divs;
}
int n, m;
vector<int> a, b;
set<pair<int, int>> segs;
void Load() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    segs.insert({a, b});
    segs.insert({b, a});
  }
}
bool RotateWorks(ll div) {
  for (const auto& seg : segs) {
    if (segs.count({(seg.first + div) % n, (seg.second + div) % n}) == 0)
      return false;
  }
  return true;
}
void Solve() {
  vector<ll> divs = GetAllDivisorsUnsorted(n);
  for (ll div : divs) {
    if (div == n) continue;
    if (RotateWorks(div)) {
      puts("Yes");
      return;
    }
  }
  puts("No");
}
int main() {
  Load();
  Solve();
}

```