#include <bits/stdc++.h>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,avx,avx2,abm,mmx")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T1, typename T2>
inline void chkmin(T1 &x, const T2 &y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
ll n, k;
struct segm {
  ll l, r, ind;
  segm() {}
};
vector<segm> a;
bool check(vector<segm> have) {
  vector<ll> pos(n + 200);
  for (auto i : have) {
    for (ll j = i.l; j <= i.r; j++) {
      pos[j]++;
    }
  }
  return (*max_element(pos.begin(), pos.end())) <= k;
}
void solve() {
  cin >> n >> k;
  a.resize(n);
  for (ll i = 0; i < n; i++) {
    cin >> a[i].l >> a[i].r;
    a[i].ind = i;
  }
  sort(a.begin(), a.end(),
       [&](segm i, segm j) { return tie(i.r, j.l) < tie(j.r, i.l); });
  vector<ll> ans;
  vector<segm> have;
  for (auto i : a) {
    have.push_back(i);
    if (!check(have)) {
      have.pop_back();
      ans.push_back(i.ind);
    }
  }
  cout << ans.size() << endl;
  for (auto i : ans) cout << i + 1 << " ";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Please create a solution in cpp to the following 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 

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,avx,avx2,abm,mmx")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T1, typename T2>
inline void chkmin(T1 &x, const T2 &y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
ll n, k;
struct segm {
  ll l, r, ind;
  segm() {}
};
vector<segm> a;
bool check(vector<segm> have) {
  vector<ll> pos(n + 200);
  for (auto i : have) {
    for (ll j = i.l; j <= i.r; j++) {
      pos[j]++;
    }
  }
  return (*max_element(pos.begin(), pos.end())) <= k;
}
void solve() {
  cin >> n >> k;
  a.resize(n);
  for (ll i = 0; i < n; i++) {
    cin >> a[i].l >> a[i].r;
    a[i].ind = i;
  }
  sort(a.begin(), a.end(),
       [&](segm i, segm j) { return tie(i.r, j.l) < tie(j.r, i.l); });
  vector<ll> ans;
  vector<segm> have;
  for (auto i : a) {
    have.push_back(i);
    if (!check(have)) {
      have.pop_back();
      ans.push_back(i.ind);
    }
  }
  cout << ans.size() << endl;
  for (auto i : ans) cout << i + 1 << " ";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll t = 1;
  while (t--) solve();
  return 0;
}

```