#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
inline int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
inline int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
inline int mul(int a, int b) { return 1ll * a * b % P; }
inline int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 202020;
int n, k;
set<pair<int, int> > b;
pair<pair<int, int>, int> a[N];
vector<int> ans;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (0); i < (n); i++)
    scanf("%d%d", &a[i].first.first, &a[i].first.second), a[i].second = i;
  sort(a, a + n);
  int p = 0;
  for (int i = (1); i < (N); i++) {
    while (p < n && a[p].first.second >= i && a[p].first.first <= i) {
      b.insert({a[p].first.second, a[p].second});
      p++;
    }
    while ((int)b.size() && b.begin()->first < i) b.erase(*b.begin());
    while ((int)b.size() > k) {
      ans.push_back(b.rbegin()->second);
      b.erase(*b.rbegin());
    }
  }
  printf("%d\n", (int)ans.size());
  for (int &id : ans) printf("%d ", id + 1);
  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>
using namespace std;
const int P = 1e9 + 7;
inline int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
inline int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
inline int mul(int a, int b) { return 1ll * a * b % P; }
inline int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 202020;
int n, k;
set<pair<int, int> > b;
pair<pair<int, int>, int> a[N];
vector<int> ans;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (0); i < (n); i++)
    scanf("%d%d", &a[i].first.first, &a[i].first.second), a[i].second = i;
  sort(a, a + n);
  int p = 0;
  for (int i = (1); i < (N); i++) {
    while (p < n && a[p].first.second >= i && a[p].first.first <= i) {
      b.insert({a[p].first.second, a[p].second});
      p++;
    }
    while ((int)b.size() && b.begin()->first < i) b.erase(*b.begin());
    while ((int)b.size() > k) {
      ans.push_back(b.rbegin()->second);
      b.erase(*b.rbegin());
    }
  }
  printf("%d\n", (int)ans.size());
  for (int &id : ans) printf("%d ", id + 1);
  return 0;
}

```