find all four sum numbers

/*
Given an array A of size N, find all combination of four elements in the array whose sum is equal to a given value K. For example, if the 
given array is {10, 2, 3, 4, 5, 9, 7, 8} and K = 23, one of the quadruple is “3 5 7 8” (3 + 5 + 7 + 8 = 23).

The output should contain only unique quadrples  For example, if input array is {1, 1, 1, 1, 1, 1} and K = 4, then output should be only one 
quadrple {1, 1, 1, 1}
 

Input:
The first line of input contains an integer T denoting the no of test cases. Then T test cases follow. Each test case contains two lines. The 
first line of input contains two integers N and K. Then in the next line are N space separated values of the array.

Output:
For each test case in a new line print all the quadruples present in the array separated by space which sums up to value of K. Each quadruple 
is unique which are separated by a delimeter "$" and are in increasing order.

Constraints:
1 <= T <= 100
1 <= N <= 100
-1000 <= K <= 1000
-100 <= A[] <= 100

Example:
Input:
2
5 3
0 0 2 1 1 
7 23
10 2 3 4 5 7 8

Output:
0 0 1 2 $
2 3 8 10 $2 4 7 10 $3 5 7 8 $
*/
#include <bits/stdc++.h>
using namespace std;
int main() {
	int t; cin>>t; while(t--){
	    int n,k; cin>>n>>k;
	    int a[n]; for(int i=0; i<n; i++) cin>>a[i];
	    sort(a,a+n);
	    vector< vector<int> > res;
	    for(int i=0; i<n; i++ ){
	        for(int j=i+1; j<n; j++){
	            int left=j+1;
	            int right=n-1;
	            while(left<right){
	                int s=a[i]+a[j]+a[left]+a[right];
	                if(s==k){
	                    vector<int> v;
	                    v.push_back(a[i]);
	                    v.push_back(a[j]);
	                    v.push_back(a[left]);
	                    v.push_back(a[right]);
	                    res.push_back(v);
	                }
	                if(s>k){
	                    do{--right;}
	                        while(a[right+1] == a[right]  && right>left);
	                    }
	                else{
	                    do{++left;}
	                    while(a[left] == a[left-1] && right>left);
	                }
	            } 
	             while(j+1<n && a[j] == a[j+1]) ++j;
	       }
	             while(i+1<n && a[i]==a[i+1]) ++i;      
	  }
	      // sort(res.begin(), res.end());
	       if(res.size()==0) cout<<-1<<endl;
	       else{
	    for(int i=0; i<res.size(); i++){
	        for(int j=0; j<res[i].size(); j++ ){
	            cout<<res[i][j]<<" "; 
	           }
	        cout<<"$";
	    }
	    cout<<endl;  
	  }
	}
	    
	}