#include <bits/stdc++.h>
namespace IO {
template <typename _T>
inline _T const& read(_T& x) {
  x = 0;
  int fh = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x *= fh;
}
inline void write(long long a) {
  if (a >= 10) write(a / 10);
  putchar(a % 10 + '0');
}
}  // namespace IO
using namespace IO;
const int maxn = 205;
int n, k, ans;
int A[maxn];
int vis[maxn];
struct Node {
  int l, r, org;
};
struct Node segments[maxn];
inline bool operator<(const Node a, const Node b) { return a.r < b.r; }
inline bool inrange(const int i, const int x) {
  return (segments[i].l <= x) && (x <= segments[i].r);
}
signed main() {
  read(n), read(k);
  for (int i = 1, l, r; i <= n; ++i) {
    read(segments[i].l), read(segments[i].r), segments[i].org = i;
    for (int j = segments[i].l; j <= segments[i].r; ++j) ++A[j];
  }
  std::sort(segments + 1, segments + 1 + n);
  for (int i = 1; i <= 200; ++i) {
    while (A[i] > k) {
      for (int j = n; j > 0; --j)
        if (inrange(j, i) && !vis[j]) {
          for (int q = segments[j].l; q <= segments[j].r; ++q) A[q]--;
          vis[j] = 1;
          ++ans;
          break;
        }
    }
  }
  write(ans);
  putchar('\n');
  for (int i = 1; i <= n; ++i)
    if (vis[i]) write(segments[i].org), putchar(' ');
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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>
namespace IO {
template <typename _T>
inline _T const& read(_T& x) {
  x = 0;
  int fh = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x *= fh;
}
inline void write(long long a) {
  if (a >= 10) write(a / 10);
  putchar(a % 10 + '0');
}
}  // namespace IO
using namespace IO;
const int maxn = 205;
int n, k, ans;
int A[maxn];
int vis[maxn];
struct Node {
  int l, r, org;
};
struct Node segments[maxn];
inline bool operator<(const Node a, const Node b) { return a.r < b.r; }
inline bool inrange(const int i, const int x) {
  return (segments[i].l <= x) && (x <= segments[i].r);
}
signed main() {
  read(n), read(k);
  for (int i = 1, l, r; i <= n; ++i) {
    read(segments[i].l), read(segments[i].r), segments[i].org = i;
    for (int j = segments[i].l; j <= segments[i].r; ++j) ++A[j];
  }
  std::sort(segments + 1, segments + 1 + n);
  for (int i = 1; i <= 200; ++i) {
    while (A[i] > k) {
      for (int j = n; j > 0; --j)
        if (inrange(j, i) && !vis[j]) {
          for (int q = segments[j].l; q <= segments[j].r; ++q) A[q]--;
          vis[j] = 1;
          ++ans;
          break;
        }
    }
  }
  write(ans);
  putchar('\n');
  for (int i = 1; i <= n; ++i)
    if (vis[i]) write(segments[i].org), putchar(' ');
  return 0;
}

```