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.*; 
import java.util.*; 
//import javafx.util.*; 
import java.math.*;
//import java.lang.*;
public class Main 
{ 
    
   // static int n;
//  static HashSet<Integer> adj[];

   // static int dist[];
   // static int remove[];
//    static boolean isLeaf[];
  //   static Long dp[][];
   // static int arr[];
  //  static boolean ok;
    // static long arr[];
    // static long sum[];
  //  static boolean ok;
    static long mod=1000000007;
 //   static HashSet<Integer> adjacency[];
    static final long oo=(long)1e18;
    public static void main(String[] args) throws IOException { 
    //    Scanner sc=new Scanner(System.in);
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        br = new BufferedReader(new InputStreamReader(System.in));
     //   int test=nextInt();
     int test=1;
        
        outer: while(test--!=0){
            
           // String s=next();
            int n=nextInt();
            int val=nextInt();
            int a1[]=new int[n];
            int a2[]=new int[n];
            int a3[]=new int[n];
            int n1=0;int n2=0;int n3=0;
            for(int i=0;i<n;i++){
                int a=nextInt();
                int t1=nextInt();
                int t2=nextInt();
                if(t1==1&&t2==1){
                    a1[n1++]=a;
                }
                else if(t1==1){
                    a2[n2++]=a;
                }
                else if(t2==1){
                    a3[n3++]=a;
                }
            }
            Arrays.sort(a1,0,n1);
            Arrays.sort(a2,0,n2);
            Arrays.sort(a3,0,n3);
            boolean ok=false;int i=0;int j=0;int k=0;long ans=0;
            while(val--!=0){
                int a=i<n1?a1[i]:100000;
                int b=j<n2?a2[j]:100000;
                int c=k<n3?a3[k]:100000;
                c=c+b;

                if(a>=100000&&c>=100000){
                    ok=true;
                    break;
                }
                else if(a<=c){
                    ans+=a;
                    i++;
                    
                }
                else{
                    ans+=c;
                    j++;k++;
                }
            }
            if(ok)
                pw.println("-1");
            else    
                pw.println(ans);

        }
           
        pw.close();
    }

    static class Pair implements Comparable<Pair>{
        double g,ai,bi,sa,sb;
        Pair(double g,double ai,double bi){
          //  this.index=index;
            this.g=g;
            this.ai=ai;
            this.bi=bi;
            sa=g/ai;
            sb=g/bi;
            //this.z=z;
        }   
        public int compareTo(Pair p){
            double a=Math.abs(sa-sb);
            double b=Math.abs(p.sa-p.sb);
            if(a>b){
                return 1;
            }
            else if(a==b){
                return 0;

            }
            else{
                return -1;
            }
        }
    }
   static void update(long tree[],int idx,long val){
        //println(idx);
        while(idx<1000006){
            
            tree[idx]+=val;
            idx+=(idx)&(-idx);
        }
    }   
    static long sum(long tree[],int idx){
        long sum=0;
        while(idx>0){
           
            sum+=tree[idx];
            idx-=(idx)&(-idx);
        }
        return sum;
    }
    public static BufferedReader br;
    public static StringTokenizer st;
    public static String next() {
        while (st == null || !st.hasMoreTokens()) {
       try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }
 
    public static Integer nextInt() {
        return Integer.parseInt(next());
    }
    public static Long nextLong() {
        return Long.parseLong(next());
    }
 
    public static Double nextDouble() {
        return Double.parseDouble(next());
    }
    // static class Pair{
    //     int x;int y;
    //     Pair(int x,int y,int z){
    //         this.x=x;
    //         this.y=y;
    //        // this.z=z;
    //      //   this.z=z;
    //       //  this.i=i;
    //     }
    // }
    // static class sorting implements Comparator<Pair>{
    //     public int compare(Pair a,Pair b){
    //         //return (a.y)-(b.y);
    //         if(a.y==b.y){
    //             return -1*(a.z-b.z);
    //         }
    //         return (a.y-b.y);
    //     }
   // }

    static long ncr(long n,long r){
        
        if(r==0)
            return 1l;
        long val=ncr(n-1,r-1);
        val=(n*val);
        val=(val/r);
        return val;
    }

    public static int[] na(int n)throws IOException{
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = nextInt();
        return a;
    }
    static class query implements Comparable<query>{
        int l,r,idx,block;
        static int len;
        query(int l,int r,int i){
            this.l=l;
            this.r=r;
            this.idx=i;
            this.block=l/len;
        }   
        public int compareTo(query a){
            return block==a.block?r-a.r:block-a.block;
        }
    }
    
    // static class sorting implements Comparator<Pair>{  
    //     public int compare(Pair a1,Pair a2){  
    //         if(o1.a==o2.a)
    //             return (o1.b>o2.b)?1:-1;  
    //         else if(o1.a>o2.a)  
    //             return 1;  
    //         else  
    //             return -1;  
    //     }
    // }  
    static boolean isPrime(int n) { 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
        if (n % 2 == 0 ||  
            n % 3 == 0) 
            return false; 
      
        for (int i = 5; 
                 i * i <= n; i = i + 6) 
            if (n % i == 0 || 
                n % (i + 2) == 0) 
                return false; 
      
        return true; 
    } 
    static int gcd(int a, int b) { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }  
    // To compute x^y under modulo m 
    static long power(long x, long y, long m){ 
        if (y == 0) 
            return 1;      
        long p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    }
    static long fast_pow(long base,long n,long M){
        if(n==0)
           return 1;
        if(n==1)
        return base;
        long halfn=fast_pow(base,n/2,M);
        if(n%2==0)
            return ( halfn * halfn ) % M;
        else
            return ( ( ( halfn * halfn ) % M ) * base ) % M;
    }
       static long modInverse(long n,long M){
        return fast_pow(n,M-2,M);
    }
    // (1,1)   

    //     (3,2)   (3,5)

}