#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-5;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
pair<int, int> p[200010];
pair<int, int> s[200010];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = (1); i <= (m); i++) {
    cin >> s[i].first >> s[i].second;
    s[i].first %= n;
    s[i].second %= n;
    if (s[i].first > s[i].second) swap(s[i].first, s[i].second);
    p[i] = s[i];
  }
  sort(s + 1, s + 1 + m);
  sort(p + 1, p + 1 + m);
  bool f = false;
  for (int i = 1; i * i <= n; i++) {
    if (n % i) continue;
    for (int j = (1); j <= (m); j++) {
      p[j].first += i;
      p[j].first %= n;
      p[j].second += i;
      p[j].second %= n;
      if (p[j].first > p[j].second) swap(p[j].first, p[j].second);
    }
    sort(p + 1, p + 1 + m);
    bool g = true;
    for (int j = (1); j <= (m); j++) {
      if (p[j] != s[j]) {
        g = false;
        break;
      }
    }
    f |= g;
    for (int j = (1); j <= (m); j++) {
      p[j].first -= i;
      p[j].first %= n;
      p[j].first += n / i;
      p[j].first %= n;
      p[j].second -= i;
      p[j].second %= n;
      p[j].second += n / i;
      p[j].second %= n;
      if (p[j].first > p[j].second) swap(p[j].first, p[j].second);
    }
    sort(p + 1, p + 1 + m);
    g = true;
    for (int j = (1); j <= (m); j++) {
      if (p[j] != s[j]) {
        g = false;
        break;
      }
    }
    if (n / i != n) f |= g;
    for (int j = (1); j <= (m); j++) {
      p[j].first -= n / i;
      p[j].first %= n;
      p[j].first += n;
      p[j].first %= n;
      p[j].second -= n / i;
      p[j].second %= n;
      p[j].second += n;
      p[j].second %= n;
    }
  }
  cout << (f ? "Yes" : "No");
  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>
#pragma warning(disable : 4996)
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-5;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
pair<int, int> p[200010];
pair<int, int> s[200010];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = (1); i <= (m); i++) {
    cin >> s[i].first >> s[i].second;
    s[i].first %= n;
    s[i].second %= n;
    if (s[i].first > s[i].second) swap(s[i].first, s[i].second);
    p[i] = s[i];
  }
  sort(s + 1, s + 1 + m);
  sort(p + 1, p + 1 + m);
  bool f = false;
  for (int i = 1; i * i <= n; i++) {
    if (n % i) continue;
    for (int j = (1); j <= (m); j++) {
      p[j].first += i;
      p[j].first %= n;
      p[j].second += i;
      p[j].second %= n;
      if (p[j].first > p[j].second) swap(p[j].first, p[j].second);
    }
    sort(p + 1, p + 1 + m);
    bool g = true;
    for (int j = (1); j <= (m); j++) {
      if (p[j] != s[j]) {
        g = false;
        break;
      }
    }
    f |= g;
    for (int j = (1); j <= (m); j++) {
      p[j].first -= i;
      p[j].first %= n;
      p[j].first += n / i;
      p[j].first %= n;
      p[j].second -= i;
      p[j].second %= n;
      p[j].second += n / i;
      p[j].second %= n;
      if (p[j].first > p[j].second) swap(p[j].first, p[j].second);
    }
    sort(p + 1, p + 1 + m);
    g = true;
    for (int j = (1); j <= (m); j++) {
      if (p[j] != s[j]) {
        g = false;
        break;
      }
    }
    if (n / i != n) f |= g;
    for (int j = (1); j <= (m); j++) {
      p[j].first -= n / i;
      p[j].first %= n;
      p[j].first += n;
      p[j].first %= n;
      p[j].second -= n / i;
      p[j].second %= n;
      p[j].second += n;
      p[j].second %= n;
    }
  }
  cout << (f ? "Yes" : "No");
  return 0;
}

```