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 
#include <bits/stdc++.h>
using namespace std;
void solve(long long tz) {
  int n, k;
  cin >> n >> k;
  int H[201] = {0};
  vector<pair<pair<int, int>, int>> arr;
  vector<pair<pair<int, int>, int>> brr;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    arr.push_back({{l, r}, i + 1});
    brr.push_back({{l, r}, i + 1});
    H[l]++;
    H[r + 1]--;
  }
  int HH[201] = {0};
  for (int i = 0; i < 201; i++) {
    if (i > 0) {
      H[i] = H[i] + H[i - 1];
    }
    HH[i] = H[i];
  }
  cout << endl;
  int cnt1 = 0;
  vector<int> ans1;
  for (int i = 0; i < 201;) {
    if (H[i] > k) {
      int maxi = -1;
      int r = i;
      int index;
      for (int j = 0; j < arr.size(); j++) {
        if (arr[j].first.first == i &&
            arr[j].first.second - arr[j].first.first > maxi) {
          maxi = arr[j].first.first - arr[j].first.first;
          r = arr[j].first.second;
          index = arr[j].second;
        }
      }
      arr.erase(arr.begin() + index - 1);
      ans1.push_back(index);
      for (int j = i; j <= r; j++) {
        H[j]--;
      }
      cnt1++;
    } else {
      i++;
    }
  }
  arr.clear();
  arr = brr;
  int cnt2 = 0;
  for (int i = 0; i < 201; i++) {
    H[i] = HH[i];
  }
  vector<int> ans2;
  for (int i = 200; i >= 0;) {
    if (H[i] > k) {
      int maxi = -1;
      int r = i;
      int index;
      for (int j = 0; j < arr.size(); j++) {
        if (arr[j].first.second == i &&
            arr[j].first.second - arr[j].first.first > maxi) {
          maxi = arr[j].first.second - arr[j].first.first;
          r = arr[j].first.first;
          index = arr[j].second;
        }
      }
      arr.erase(arr.begin() + index - 1);
      ans2.push_back(index);
      for (int j = i; j >= r; j--) {
        H[j]--;
      }
      cnt2++;
    } else {
      i--;
    }
  }
  if (cnt1 > cnt2) {
    cout << cnt2 << endl;
    for (int i = 0; i < ans2.size(); i++) {
      cout << ans2[i] << " ";
    }
  } else {
    cout << cnt1 << endl;
    for (int i = 0; i < ans1.size(); i++) {
      cout << ans1[i] << " ";
    }
  }
}
int main() {
  long long t;
  t = 1;
  long long z = 0;
  while (t--) {
    z++;
    solve(z);
  }
}
