#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 50;
struct Fenwick_Tree {
  int bit[maxn][2];
  int n;
  void init(int _n) {
    n = _n;
    memset(bit, 0, sizeof bit);
  }
  int lowbit(int x) { return x & (-x); }
  void insert(int x, int v) {
    for (int i = x; i <= n; i += lowbit(i)) {
      bit[i][0] += v;
      bit[i][1] += v * (x - 1);
    }
  }
  void upd(int l, int r, int v) {
    insert(l, v);
    insert(r + 1, -v);
  }
  int get(int x) {
    int res = 0;
    for (int i = x; i; i -= lowbit(i)) {
      res += x * bit[i][0] - bit[i][1];
    }
    return res;
  }
  int ask(int l, int r) { return get(r) - get(l - 1); }
} tr;
vector<pair<int, int> > mp[maxn];
vector<int> ans;
set<pair<int, int> > res;
int bit[maxn];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    mp[l].push_back({r, i});
    bit[l]++;
    bit[r + 1]--;
  }
  for (int i = 1; i <= maxn - 50; i++) bit[i] += bit[i - 1];
  tr.init(maxn - 50);
  for (int i = 1; i <= maxn - 50; i++) {
    int now = bit[i] - tr.ask(i, i);
    for (auto x : mp[i]) res.insert(x);
    if (now <= k) continue;
    for (int j = 0; j < now - k; j++) {
      tr.upd(i, (*res.rbegin()).first, 1);
      ans.push_back((*res.rbegin()).second);
      res.erase(res.find(*res.rbegin()));
    }
  }
  printf("%d\n", (int)ans.size());
  sort(begin(ans), end(ans));
  for (auto x : ans) printf("%d ", x);
  return 0;
}

### Prompt

Create a solution in cpp for 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>
using namespace std;
const int maxn = 2e5 + 50;
struct Fenwick_Tree {
  int bit[maxn][2];
  int n;
  void init(int _n) {
    n = _n;
    memset(bit, 0, sizeof bit);
  }
  int lowbit(int x) { return x & (-x); }
  void insert(int x, int v) {
    for (int i = x; i <= n; i += lowbit(i)) {
      bit[i][0] += v;
      bit[i][1] += v * (x - 1);
    }
  }
  void upd(int l, int r, int v) {
    insert(l, v);
    insert(r + 1, -v);
  }
  int get(int x) {
    int res = 0;
    for (int i = x; i; i -= lowbit(i)) {
      res += x * bit[i][0] - bit[i][1];
    }
    return res;
  }
  int ask(int l, int r) { return get(r) - get(l - 1); }
} tr;
vector<pair<int, int> > mp[maxn];
vector<int> ans;
set<pair<int, int> > res;
int bit[maxn];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    mp[l].push_back({r, i});
    bit[l]++;
    bit[r + 1]--;
  }
  for (int i = 1; i <= maxn - 50; i++) bit[i] += bit[i - 1];
  tr.init(maxn - 50);
  for (int i = 1; i <= maxn - 50; i++) {
    int now = bit[i] - tr.ask(i, i);
    for (auto x : mp[i]) res.insert(x);
    if (now <= k) continue;
    for (int j = 0; j < now - k; j++) {
      tr.upd(i, (*res.rbegin()).first, 1);
      ans.push_back((*res.rbegin()).second);
      res.erase(res.find(*res.rbegin()));
    }
  }
  printf("%d\n", (int)ans.size());
  sort(begin(ans), end(ans));
  for (auto x : ans) printf("%d ", x);
  return 0;
}

```