#include <bits/stdc++.h>
using namespace std;
constexpr int MOD = 1e9 + 7;
constexpr int MAXN = 2e6 + 10;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
ll N, M, K, T;
set<Pii> ss;
void AddSeg(int u, int v) {
  int a = (v - u + N) % N, b = (u - v + N) % N;
  if (a < b)
    ss.emplace(u, v);
  else
    ss.emplace(v, u);
}
bool Check(int k) {
  for (auto &p : ss) {
    int a = (p.first + k) % N, b = (p.second + k) % N;
    auto it0 = ss.find(make_pair(a, b));
    auto it1 = ss.find(make_pair(b, a));
    if (it0 == ss.end() && it1 == ss.end()) return false;
  }
  return true;
}
bool Solution() {
  for (int d = 2; d <= sqrt(N); ++d) {
    if (N % d == 0) {
      if (Check(d) || Check(N / d)) {
        return true;
      }
    }
  }
  return Check(1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  while (cin >> N >> M) {
    int u, v;
    for (int i = (int)(0); i < (int)(M); ++i) {
      cin >> u >> v;
      AddSeg(u - 1, v - 1);
    }
    if (Solution())
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  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>
using namespace std;
constexpr int MOD = 1e9 + 7;
constexpr int MAXN = 2e6 + 10;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
ll N, M, K, T;
set<Pii> ss;
void AddSeg(int u, int v) {
  int a = (v - u + N) % N, b = (u - v + N) % N;
  if (a < b)
    ss.emplace(u, v);
  else
    ss.emplace(v, u);
}
bool Check(int k) {
  for (auto &p : ss) {
    int a = (p.first + k) % N, b = (p.second + k) % N;
    auto it0 = ss.find(make_pair(a, b));
    auto it1 = ss.find(make_pair(b, a));
    if (it0 == ss.end() && it1 == ss.end()) return false;
  }
  return true;
}
bool Solution() {
  for (int d = 2; d <= sqrt(N); ++d) {
    if (N % d == 0) {
      if (Check(d) || Check(N / d)) {
        return true;
      }
    }
  }
  return Check(1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  while (cin >> N >> M) {
    int u, v;
    for (int i = (int)(0); i < (int)(M); ++i) {
      cin >> u >> v;
      AddSeg(u - 1, v - 1);
    }
    if (Solution())
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}

```