#include <bits/stdc++.h>
using namespace std;
long long mult(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  if (a >= p) a %= p;
  if (b >= p) b %= p;
  if (a + b > p)
    return (a + b) % p;
  else
    return a + b;
}
long long sub(long long a, long long b, long long p = 1000000007) {
  return (a % p - b % p + p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = mult(r, n, p);
    n = mult(n, n, p);
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
vector<pair<long long, long long> > v, v2;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    if (x > y) swap(x, y);
    v.push_back({x, y});
  }
  sort((v).begin(), (v).end());
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      v2.clear();
      for (auto j : v) {
        long long x = j.first + i, y = j.second + i;
        if (x > n) x -= n;
        if (y > n) y -= n;
        if (x > y) swap(x, y);
        v2.push_back({x, y});
      }
      sort((v2).begin(), (v2).end());
      if (v == v2) {
        cout << "YES\n";
        return 0;
      }
    }
  }
  cout << "NO";
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
long long mult(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  if (a >= p) a %= p;
  if (b >= p) b %= p;
  if (a + b > p)
    return (a + b) % p;
  else
    return a + b;
}
long long sub(long long a, long long b, long long p = 1000000007) {
  return (a % p - b % p + p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = mult(r, n, p);
    n = mult(n, n, p);
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
vector<pair<long long, long long> > v, v2;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    if (x > y) swap(x, y);
    v.push_back({x, y});
  }
  sort((v).begin(), (v).end());
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      v2.clear();
      for (auto j : v) {
        long long x = j.first + i, y = j.second + i;
        if (x > n) x -= n;
        if (y > n) y -= n;
        if (x > y) swap(x, y);
        v2.push_back({x, y});
      }
      sort((v2).begin(), (v2).end());
      if (v == v2) {
        cout << "YES\n";
        return 0;
      }
    }
  }
  cout << "NO";
}

```