Problem: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.

Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.

There are n books in the family library. The i-th book is described by three integers: t_i — the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).

So they need to choose some books from the given n books in such a way that:

  * Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set; 
  * the total reading time of these books is minimized (they are children and want to play and joy as soon a possible). 



The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.

Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5).

The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 ≤ t_i ≤ 10^4, 0 ≤ a_i, b_i ≤ 1), where:

  * t_i — the amount of time required for reading the i-th book; 
  * a_i equals 1 if Alice likes the i-th book and 0 otherwise; 
  * b_i equals 1 if Bob likes the i-th book and 0 otherwise. 

Output

If there is no solution, print only one integer -1. Otherwise print one integer T — the minimum total reading time of the suitable set of books.

Examples

Input

8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0


Output

18


Input

5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0


Output

8


Input

5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1


Output

-1

Solution (in JAVA):

Solution:
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;
 
public class Solution 
{ 
    
//     public static int gcd(int a,int b){
//         if(a==1 ||b==1)
//         return 1;
//         if(b==0)
//         return a;
//         return gcd(b, a%b);
//     }
//       public static HashMap<Integer,Integer> primeFactors(int n) 
//     { 
//       HashMap<Integer,Integer>  h=new  HashMap<>  ();
//         while (n%2==0) 
//         { 
//             if(h.containsKey(2))
//          h.add(2,h.get(2)+1);
//     else
//     h.put(2,1);
//             n /= 2; 
//         } 
  
//         // n must be odd at this point.  So we can 
//         // skip one element (Note i = i +2) 
//         for (int i = 3; i <= Math.sqrt(n); i+= 2) 
//         { 
//             // While i divides n, print i and divide n 
//             while (n%i == 0) 
//             { 
//                     if(h.containsKey(i))
//          h.add(i,h.get(i)+1);
//     else
//     h.put(i,1);
//                 n /= i; 
//             } 
//         } 
  
//         // This condition is to handle the case whien 
//         // n is a prime number greater than 2 
//         if (n > 2) {
//               if(h.containsKey(n))
//          h.add(n,h.get(n)+1);
//     else
//     h.put(n,1);
         
//         }
//         return h;
//     } 
//   public  static boolean isPowerOfTwo (int x) 
//     { 
//       /* First x in the below expression is  
//         for the case when x is 0 */
//         return x!=0 && ((x&(x-1)) == 0); 
          
//     } 
    public static void main(String[] args) throws IOException 
    { 
  
       BufferedReader br = new BufferedReader( 
                             new InputStreamReader(System.in));
        StringBuilder out=new StringBuilder(); 
        StringTokenizer st=new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken());
        int k=Integer.parseInt(st.nextToken());
       int t=n;
       ArrayList<Integer> bo=new ArrayList<>();
         ArrayList<Integer> aa=new ArrayList<>();
           ArrayList<Integer> bb=new ArrayList<>();
        while(t-->0){
        st=new StringTokenizer(br.readLine());
     int s=Integer.parseInt(st.nextToken());
        int a=Integer.parseInt(st.nextToken());
      int b=Integer.parseInt(st.nextToken());
      if(a==1 && b==1)
      bo.add(s);
      else if(a==0 && b==1)
      bb.add(s);
      else if(b==0 && a==1)
      aa.add(s);
        }
       
        if(aa.size()+bo.size()<k || bb.size()+bo.size()<k)
        {
            out.append(-1);
            
        }else{
            long ans=0;
             int i=0,j=0,l=0;
             int s1=k,s2=k;
             Collections.sort(bo);
              Collections.sort(bb);
               Collections.sort(aa);
             while(i<bo.size() ||j<aa.size()||l<bb.size()){
                 if(s1==0 && s2==0)
                 break;
                 long a1,a2,a3;
                 if(i>=bo.size())
                 a1=Integer.MAX_VALUE;
                 else
                 a1=bo.get(i);
                  if(j>=aa.size())
                 a2=Integer.MAX_VALUE;
                 else
                 a2=aa.get(j);
                  if(l>=bb.size())
                 a3=Integer.MAX_VALUE;
                 else
                 a3=bb.get(l);
                 if(a1<=a2+a3){
                     s1--;s2--;
                     ans=ans+a1;
                     i++;
                 }else{
                     s1--;s2--;
                     ans=ans+a2+a3;
                     j++;l++;
                 }
              //   System.out.println(a1+"*"+a2+"*"+a3+"/"+ans);
             }
             out.append(ans);
        }
       
System.out.println(out);


}
}
// class Graph 
// { 
//     private int V;   // No. of vertices 
  
//     // Array  of lists for Adjacency List Representation 
//      ArrayList<pair> adj[]; 
  
//     // Constructor 
//     Graph(int v) 
//     { 
//         V = v; 
//         adj = new ArrayList[v]; 
//         for (int i=0; i<v; ++i) 
//             adj[i] = new ArrayList(); 
//     } 
  
//     //Function to add an edge into the graph 
//     void addEdge(int v, int w,int x) 
//     { 
//         adj[v].add(new pair(w,x));
//         adj[w].add(new pair(v,x));// Add w to v's list. 
//     } 
  
//     // A function used by DFS 
// }
// class pair{
//     int x;
//     int y;
//     public pair(int xx,int yy){
//         x=xx;
//         y=yy;
        
//     }
// }
class SortT implements Comparator<String> 
{ 
   
    public int compare(String a, String b) 
    { 
       
        return a.length() - b.length(); 
    } 
}