#include <bits/stdc++.h>
using namespace std;
inline int in() {
  int x;
  scanf("%d", &x);
  return x;
}
bool Check(int N, int pos) { return (bool)(N & (1 << pos)); }
int Set(int N, int pos) { return N = N | (1 << pos); }
int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1};
int hr[] = {-2, -2, -1, +1, +2, +2, -1, +1};
int hc[] = {+1, -1, +2, +2, -1, +1, -2, -2};
int dx[] = {+1, -1, +0, +0};
int dy[] = {+0, +0, +1, -1};
const int MAXN = (int)1e5 + 9;
struct node {
  int l, r, i;
  node() {}
  node(int l1, int r1, int i1) {
    l = l1, r = r1;
    i = i1;
  }
  bool operator<(const node &nd) const {
    if (l == nd.l) {
      return r < nd.r;
    } else
      return l < nd.l;
  }
};
int main() {
  int i, j, k, l, m, n, p, q, x, y, u, v, r, tc, t;
  cin >> n >> k;
  vector<node> vv;
  for (i = 1; i <= n; i++) {
    cin >> l >> r;
    vv.push_back(node(l, r, i));
  }
  sort(vv.begin(), vv.end());
  int ara[210] = {0};
  vector<int> ans, ans2;
  for (i = 0; i < n; i++) {
    l = vv[i].l, r = vv[i].r;
    for (j = l; j <= r; j++) ara[j]++;
    bool flag = false;
    for (j = 1; j <= 200; j++) {
      if (ara[j] > k) {
        flag = true;
        break;
      }
    }
    if (flag) {
      ans.push_back(vv[i].i);
      for (j = l; j <= r; j++) ara[j]--;
    }
  }
  memset(ara, 0, sizeof(ara));
  for (i = n - 1; i >= 0; i--) {
    l = vv[i].l, r = vv[i].r;
    for (j = l; j <= r; j++) ara[j]++;
    bool flag = false;
    for (j = 1; j <= 200; j++) {
      if (ara[j] > k) {
        flag = true;
        break;
      }
    }
    if (flag) {
      ans2.push_back(vv[i].i);
      for (j = l; j <= r; j++) ara[j]--;
    }
  }
  if (ans2.size() < ans.size()) ans = ans2;
  cout << ans.size() << endl;
  for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}

### Prompt

Generate 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>
using namespace std;
inline int in() {
  int x;
  scanf("%d", &x);
  return x;
}
bool Check(int N, int pos) { return (bool)(N & (1 << pos)); }
int Set(int N, int pos) { return N = N | (1 << pos); }
int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1};
int hr[] = {-2, -2, -1, +1, +2, +2, -1, +1};
int hc[] = {+1, -1, +2, +2, -1, +1, -2, -2};
int dx[] = {+1, -1, +0, +0};
int dy[] = {+0, +0, +1, -1};
const int MAXN = (int)1e5 + 9;
struct node {
  int l, r, i;
  node() {}
  node(int l1, int r1, int i1) {
    l = l1, r = r1;
    i = i1;
  }
  bool operator<(const node &nd) const {
    if (l == nd.l) {
      return r < nd.r;
    } else
      return l < nd.l;
  }
};
int main() {
  int i, j, k, l, m, n, p, q, x, y, u, v, r, tc, t;
  cin >> n >> k;
  vector<node> vv;
  for (i = 1; i <= n; i++) {
    cin >> l >> r;
    vv.push_back(node(l, r, i));
  }
  sort(vv.begin(), vv.end());
  int ara[210] = {0};
  vector<int> ans, ans2;
  for (i = 0; i < n; i++) {
    l = vv[i].l, r = vv[i].r;
    for (j = l; j <= r; j++) ara[j]++;
    bool flag = false;
    for (j = 1; j <= 200; j++) {
      if (ara[j] > k) {
        flag = true;
        break;
      }
    }
    if (flag) {
      ans.push_back(vv[i].i);
      for (j = l; j <= r; j++) ara[j]--;
    }
  }
  memset(ara, 0, sizeof(ara));
  for (i = n - 1; i >= 0; i--) {
    l = vv[i].l, r = vv[i].r;
    for (j = l; j <= r; j++) ara[j]++;
    bool flag = false;
    for (j = 1; j <= 200; j++) {
      if (ara[j] > k) {
        flag = true;
        break;
      }
    }
    if (flag) {
      ans2.push_back(vv[i].i);
      for (j = l; j <= r; j++) ara[j]--;
    }
  }
  if (ans2.size() < ans.size()) ans = ans2;
  cout << ans.size() << endl;
  for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}

```