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;
vector<pair<pair<long long int, long long int>, long long int> > v;
vector<long long int> res;
long long int tree[200005 * 4], lazy[200005 * 4];
long long int query_tree(int node, int a, int b, int i, int j) {
  int left = node << 1;
  int right = (node << 1) + 1;
  int mid = (a + b) / 2;
  if (lazy[node] != 0) {
    tree[node] += lazy[node];
    if (a != b) {
      lazy[left] += lazy[node];
      lazy[right] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (a > b || a > j || b < i) return 0;
  if (a >= i && b <= j) return tree[node];
  long long int q1 = query_tree(left, a, mid, i, j);
  long long int q2 = query_tree(right, mid + 1, b, i, j);
  return max(q1, q2);
}
void update_tree(int node, int a, int b, int i, int j, long long int newValue) {
  int left = node << 1;
  int right = (node << 1) + 1;
  int mid = (a + b) / 2;
  if (lazy[node] != 0) {
    tree[node] += lazy[node];
    if (a != b) {
      lazy[left] += lazy[node];
      lazy[right] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (a > b || a > j || b < i) return;
  if (a >= i && b <= j) {
    tree[node] += newValue;
    if (a != b) {
      lazy[left] += newValue;
      lazy[right] += newValue;
    }
    return;
  }
  update_tree(left, a, mid, i, j, newValue);
  update_tree(right, mid + 1, b, i, j, newValue);
  tree[node] = max(tree[left], tree[right]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i, j, k, n, li, ri;
  scanf("%lld%lld", &n, &k);
  for (i = 1; i <= n; i++) {
    scanf("%lld%lld", &li, &ri);
    v.push_back(pair<pair<long long int, long long int>, long long int>(
        pair<long long int, long long int>(ri, li * (-1)), i));
  }
  sort(v.begin(), v.end());
  for (i = 0; i < n; i++) {
    long long int p =
        query_tree(1, 1, 200005, v[i].first.second * (-1), v[i].first.first);
    if (p < k) {
      update_tree(1, 1, 200005, v[i].first.second * (-1), v[i].first.first,
                  1ll);
    } else {
      res.push_back(v[i].second);
    }
  }
  long long int x = res.size();
  printf("%lld\n", x);
  for (i = 0; i < x; i++) {
    printf("%lld ", res[i]);
  }
  return 0;
}