Problem: The only difference between easy and hard versions is constraints.

You are given n segments on the coordinate axis OX. Segments can intersect, lie inside each other and even coincide. The i-th segment is [l_i; r_i] (l_i ≤ r_i) and it covers all integer points j such that l_i ≤ j ≤ r_i.

The integer point is called bad if it is covered by strictly more than k segments.

Your task is to remove the minimum number of segments so that there are no bad points at all.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 200) — the number of segments and the maximum number of segments by which each integer point can be covered.

The next n lines contain segments. The i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 200) — the endpoints of the i-th segment.

Output

In the first line print one integer m (0 ≤ m ≤ n) — the minimum number of segments you need to remove so that there are no bad points.

In the second line print m distinct integers p_1, p_2, ..., p_m (1 ≤ p_i ≤ n) — indices of segments you remove in any order. If there are multiple answers, you can print any of them.

Examples

Input


7 2
11 11
9 11
7 8
8 9
7 8
9 11
7 9


Output


3
1 4 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 2 4 


Input


6 1
2 3
3 3
2 3
2 2
2 3
2 3


Output


4
1 3 5 6 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  cin >> x;
}
template <typename T, typename T0>
void read(T &x, T0 &y) {
  cin >> x >> y;
}
template <typename T, typename T0, typename T1>
void read(T &x, T0 &y, T1 &z) {
  cin >> x >> y >> z;
}
template <typename T, typename T0, typename T1, typename T2>
void read(T &x, T0 &y, T1 &z, T2 &w) {
  cin >> x >> y >> z >> w;
}
template <typename T, typename T0>
void read(pair<T, T0> &p) {
  cin >> p.first >> p.second;
}
template <typename T>
void read(vector<T> &oneD) {
  for (long long i = 0; i < oneD.size(); i++) {
    read(oneD[i]);
  }
}
template <typename T>
void read(T oneD[], int n) {
  for (long long i = 0; i < n; i++) {
    read(oneD[i]);
  }
}
template <typename T>
void write(T &x) {
  cout << x << " ";
}
template <typename T, typename T0>
void write(T &x, T0 &y) {
  cout << x << " " << y << "\n";
}
template <typename T, typename T0, typename T1>
void write(T &x, T0 &y, T1 &z) {
  cout << x << " " << y << " " << z << "\n";
}
template <typename T, typename T0, typename T1, typename T2>
void write(T &x, T0 &y, T1 &z, T2 &w) {
  cout << x << " " << y << " " << z << " " << w << "\n";
}
template <typename T, typename T0>
void write(pair<T, T0> &p) {
  write(p.first);
  write(p.second);
  cout << "\n";
}
template <typename T>
void write(vector<T> &oneD) {
  for (long long i = 0; i < oneD.size(); i++) {
    write(oneD[i]);
  }
  cout << "\n";
}
template <typename T>
void write(T oneD[], int n) {
  for (long long i = 0; i < n; i++) {
    write(oneD[i]);
  }
  cout << "\n";
}
template <typename T, typename T0>
void write(map<T, T0> &mpp) {
  for (auto it : mpp) {
    write(it.first);
    cout << ": ";
    write(it.second);
    cout << "\n";
  }
  cout << "\n";
}
vector<long long> sieve;
void Sieve(int N) {
  const long long maxn = N;
  sieve.resize(maxn);
  for (long long(i) = 0; (i) < (maxn); ++(i)) sieve[i] = i;
  sieve[1] = -1;
  sieve[0] = -1;
  for (long long(i) = 2; (i) <= (maxn); (i) += (1))
    if (i == sieve[i])
      for (long long j = 2 * i; j < maxn; j += i)
        if (sieve[j] == j) sieve[j] = i;
}
long long extended_GCD(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long power(long long a, long long b, long long m = 1000000007) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long modinv(long long a, long long mod = 1000000007) {
  long long x, y;
  extended_GCD(a, mod, x, y);
  if (x < 0) x += mod;
  return x;
}
const int N = 1e5 + 1;
void solve() {
  int n, k;
  cin >> n >> k;
  map<int, vector<pair<long long, long long> > > L;
  for (long long(i) = 0; (i) < (n); ++(i)) {
    int l, r;
    cin >> l >> r;
    L[l].push_back({r, i});
  }
  vector<long long> ans;
  multiset<pair<long long, long long> > R;
  for (int i = 1; i < N; ++i) {
    for (auto &v : L[i]) {
      R.insert(v);
    }
    while (R.size() > k) {
      auto x = --R.end();
      ans.push_back(x->second);
      R.erase(x);
    }
    while (R.size() > 0 && R.begin()->first <= i) {
      R.erase(R.begin());
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << "\n";
  for (long long(i) = 0; (i) < (ans.size()); ++(i)) {
    cout << ans[i] + 1 << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}