#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000 + 5;
struct node {
  int l, r;
  int idx;
  bool operator<(const node& rhs) const {
    if (r != rhs.r) return r < rhs.r;
    if (l != rhs.l) return l < rhs.l;
    return idx < rhs.idx;
  }
} in[maxn];
int cnt[maxn];
bool cmp(const node& a, const node& b) { return a.l < b.l; }
int main(void) {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &in[i].l, &in[i].r);
    in[i].idx = i;
    cnt[in[i].l]++;
    cnt[in[i].r + 1]--;
  }
  sort(in + 1, in + n + 1, cmp);
  int pos = 1;
  priority_queue<node> pq;
  vector<int> res;
  for (int i = 1; i < maxn; i++) {
    while (pos <= n && in[pos].l == i) {
      pq.push(in[pos++]);
    }
    cnt[i] += cnt[i - 1];
    while (cnt[i] > k) {
      cnt[i]--;
      cnt[pq.top().r + 1]++;
      res.push_back(pq.top().idx);
      pq.pop();
    }
  }
  printf("%d\n", res.size());
  for (int x : res) printf("%d ", x);
  printf("\n");
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
using namespace std;
const int maxn = 200000 + 5;
struct node {
  int l, r;
  int idx;
  bool operator<(const node& rhs) const {
    if (r != rhs.r) return r < rhs.r;
    if (l != rhs.l) return l < rhs.l;
    return idx < rhs.idx;
  }
} in[maxn];
int cnt[maxn];
bool cmp(const node& a, const node& b) { return a.l < b.l; }
int main(void) {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &in[i].l, &in[i].r);
    in[i].idx = i;
    cnt[in[i].l]++;
    cnt[in[i].r + 1]--;
  }
  sort(in + 1, in + n + 1, cmp);
  int pos = 1;
  priority_queue<node> pq;
  vector<int> res;
  for (int i = 1; i < maxn; i++) {
    while (pos <= n && in[pos].l == i) {
      pq.push(in[pos++]);
    }
    cnt[i] += cnt[i - 1];
    while (cnt[i] > k) {
      cnt[i]--;
      cnt[pq.top().r + 1]++;
      res.push_back(pq.top().idx);
      pq.pop();
    }
  }
  printf("%d\n", res.size());
  for (int x : res) printf("%d ", x);
  printf("\n");
  return 0;
}

```