#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const long double pai = acos(-1);
int n, m;
int x[200009], y[200009];
set<pair<int, int> > s;
pair<int, int> edit(int x, int y) {
  if (x > y) swap(x, y);
  return {x, y};
}
void fill() {
  for (int i = 0; i < m; i++) s.insert(edit(x[i], y[i]));
}
pair<int, int> find(int id, int l) {
  int X = (x[id] + l) % n;
  int Y = (y[id] + l) % n;
  return {X, Y};
}
int check(int l) {
  fill();
  for (int i = 0; i < m; i++) {
    pair<int, int> x = find(i, l);
    x = edit(x.first, x.second);
    if (s.find(x) == s.end()) return 0;
    s.erase(x);
  }
  return 1;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
    x[i]--, y[i]--;
    if (x[i] > y[i]) swap(x[i], y[i]);
  }
  vector<int> d;
  for (int i = 1; i * i <= n; i++) {
    if (n % i) continue;
    d.push_back(i);
    if (i * i == n || n / i == n) continue;
    d.push_back(n / i);
  }
  for (auto u : d)
    if (check(u)) cout << "YES" << endl, exit(0);
  cout << "NO" << endl;
}

### Prompt

In Cpp, your task is to solve 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 long long inf = 1e18;
const long double pai = acos(-1);
int n, m;
int x[200009], y[200009];
set<pair<int, int> > s;
pair<int, int> edit(int x, int y) {
  if (x > y) swap(x, y);
  return {x, y};
}
void fill() {
  for (int i = 0; i < m; i++) s.insert(edit(x[i], y[i]));
}
pair<int, int> find(int id, int l) {
  int X = (x[id] + l) % n;
  int Y = (y[id] + l) % n;
  return {X, Y};
}
int check(int l) {
  fill();
  for (int i = 0; i < m; i++) {
    pair<int, int> x = find(i, l);
    x = edit(x.first, x.second);
    if (s.find(x) == s.end()) return 0;
    s.erase(x);
  }
  return 1;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
    x[i]--, y[i]--;
    if (x[i] > y[i]) swap(x[i], y[i]);
  }
  vector<int> d;
  for (int i = 1; i * i <= n; i++) {
    if (n % i) continue;
    d.push_back(i);
    if (i * i == n || n / i == n) continue;
    d.push_back(n / i);
  }
  for (auto u : d)
    if (check(u)) cout << "YES" << endl, exit(0);
  cout << "NO" << endl;
}

```