#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 5;
const long long INF = 1e9 + 7;
template <typename T>
inline void ReadInt(T &x) {
  char c;
  while (c = getchar(), c < '0' || c > '9')
    ;
  x = c - '0';
  while (c = getchar(), c >= '0' && c <= '9')
    x = ((x << 3) + (x << 1)) + c - '0';
}
template <typename T>
void WriteInt(T x) {
  if (x > 9) WriteInt(x / 10);
  putchar(x % 10 + '0');
}
struct dl {
  long long l, r, vt;
} a[N];
long long n, k;
multiset<pair<long long, long long> > S;
vector<long long> kq;
bool cmp(const dl &a, const dl &b) {
  if (a.l != b.l) return a.l < b.l;
  if (a.r != b.r) return a.r < b.r;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    long long x, y;
    cin >> x >> y;
    a[i] = {x, y};
    a[i].vt = i;
  }
  sort(a + 1, a + n + 1, cmp);
  long long t = 0;
  for (long long i = 1; i <= n; i++) t = max(t, a[i].r);
  long long it = 0;
  for (long long i = 1; i <= t; i++) {
    while (it + 1 <= n && a[it + 1].l <= i) {
      it++;
      S.insert(pair<long long, long long>(a[it].r, a[it].vt));
    }
    while (S.size() > k) {
      pair<long long, long long> tt = *S.rbegin();
      S.erase(S.find(*S.rbegin()));
      kq.push_back(tt.second);
    }
    while (not S.empty() && (*S.begin()).first <= i) S.erase(S.begin());
  }
  sort(kq.begin(), kq.end());
  cout << size_t(kq.size()) << endl;
  for (auto x : kq) {
    cout << 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 long long N = 1e6 + 5;
const long long INF = 1e9 + 7;
template <typename T>
inline void ReadInt(T &x) {
  char c;
  while (c = getchar(), c < '0' || c > '9')
    ;
  x = c - '0';
  while (c = getchar(), c >= '0' && c <= '9')
    x = ((x << 3) + (x << 1)) + c - '0';
}
template <typename T>
void WriteInt(T x) {
  if (x > 9) WriteInt(x / 10);
  putchar(x % 10 + '0');
}
struct dl {
  long long l, r, vt;
} a[N];
long long n, k;
multiset<pair<long long, long long> > S;
vector<long long> kq;
bool cmp(const dl &a, const dl &b) {
  if (a.l != b.l) return a.l < b.l;
  if (a.r != b.r) return a.r < b.r;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    long long x, y;
    cin >> x >> y;
    a[i] = {x, y};
    a[i].vt = i;
  }
  sort(a + 1, a + n + 1, cmp);
  long long t = 0;
  for (long long i = 1; i <= n; i++) t = max(t, a[i].r);
  long long it = 0;
  for (long long i = 1; i <= t; i++) {
    while (it + 1 <= n && a[it + 1].l <= i) {
      it++;
      S.insert(pair<long long, long long>(a[it].r, a[it].vt));
    }
    while (S.size() > k) {
      pair<long long, long long> tt = *S.rbegin();
      S.erase(S.find(*S.rbegin()));
      kq.push_back(tt.second);
    }
    while (not S.empty() && (*S.begin()).first <= i) S.erase(S.begin());
  }
  sort(kq.begin(), kq.end());
  cout << size_t(kq.size()) << endl;
  for (auto x : kq) {
    cout << x << ' ';
  }
  return 0;
}

```