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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct segmentTree {
  int low, high;
  segmentTree *left;
  segmentTree *right;
  int val, lazy;
  void update_val() { val = left->val + right->val; }
  segmentTree(int l, int h, int default_val) {
    low = l;
    high = h;
    lazy = 0;
    if (l < h) {
      int mid = (low + high) / 2;
      left = (segmentTree *)malloc(sizeof(segmentTree));
      right = (segmentTree *)malloc(sizeof(segmentTree));
      *left = segmentTree(low, mid, default_val);
      *right = segmentTree(mid + 1, high, default_val);
      update_val();
    } else {
      val = default_val;
    }
  }
  segmentTree(int l, int h, int *arr) {
    low = l;
    high = h;
    lazy = 0;
    if (l < h) {
      int mid = (low + high) / 2;
      left = (segmentTree *)malloc(sizeof(segmentTree));
      right = (segmentTree *)malloc(sizeof(segmentTree));
      *left = segmentTree(low, mid, arr);
      *right = segmentTree(mid + 1, high, arr);
      update_val();
    } else {
      val = arr[l];
    }
  }
  void push_down() {
    val += lazy * (high - low + 1);
    if (low < high) {
      left->lazy += lazy;
      right->lazy += lazy;
    }
    lazy = 0;
  }
  int get() {
    push_down();
    return val;
  }
  int get_range(int l, int h) {
    push_down();
    if (low > h || high < l) {
      return 0;
    } else if (low >= l && high <= h) {
      return val;
    } else {
      return left->get_range(l, h) + right->get_range(l, h);
    }
  }
  void update_range(int l, int h, int diff) {
    push_down();
    if (low > h || high < l) {
      return;
    }
    if (low >= l && high <= h) {
      lazy += diff;
      push_down();
    } else {
      left->update_range(l, h, diff);
      right->update_range(l, h, diff);
      update_val();
    }
  }
  void change_val(int k, int vl) {
    push_down();
    if (low > k || high < k) {
      return;
    } else if (low == high) {
      val = vl;
    } else {
      left->change_val(k, vl);
      right->change_val(k, vl);
      update_val();
    }
  }
};
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  segmentTree st = segmentTree(0, 200001, 0);
  vector<tuple<int, int, int>> segs;
  set<tuple<int, int, int>> s;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    st.update_range(l, r, 1);
    segs.push_back({l, r, i});
  }
  sort(segs.begin(), segs.end());
  vector<int> ans;
  int j = 0;
  for (int i = 1; i <= 200000; i++) {
    while (j < n && get<0>(segs[j]) <= i) {
      s.insert({get<1>(segs[j]), get<0>(segs[j]), get<2>(segs[j])});
      j++;
    }
    while (st.get_range(i, i) > k) {
      tuple<int, int, int> tp = *prev(s.end());
      s.erase(prev(s.end()));
      int l = get<1>(tp), r = get<0>(tp);
      ans.push_back(get<2>(tp));
      st.update_range(l, r, -1);
    }
  }
  cout << (int)ans.size() << "\n";
  for (auto ele : ans) {
    cout << ele + 1 << " ";
  }
  cout << "\n";
  return 0;
}