Rotten Ropes

Suppose we have n ropes of equal length and we want to use them to lift some heavy object. A tear-off weight t is associated to each rope, that is, if we try to lift an object, heavier than t with that rope, it will tear off. But we can fasten a number of ropes to the heavy object (in parallel), and lift it with all the fastened ropes. When using k ropes to lift a heavy object with weight w, we assume that each of the k ropes, regardless of its tear-off weight, is responsible for lifting a weight of w/k. However, if w/k > t for some rope with tear-off weight of t, that rope will tear off. For example, three ropes with tear-off weights of 1, 10, and 15, when all three are fastened to an object, can not lift an object with weight more than 3, unless the weaker one tears-off. But the second rope, may lift by itself, an object with weight at most 10. Given the tear-off weights of n ropes, your task is to find the weight of the heaviest object that can be lifted by fastening a subset of the given ropes without any of them tearing off.

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 ≤ n ≤ 1000) which is the number of ropes. Following the first line, there is a single line containing n integers between 1 and 10000 which are the tear-off weights of the ropes, separated by blank characters.

Output

Each line of the output file should contain a single number, which is the largest weight that can be lifted in the corresponding test case without tearing off any rope chosen.

Sample Input

2
3
10 1 15
2
10 15

Sample Output

20
20

Source: Asia - Tehran 2003



my code:


#include <iostream>
#include <string>
using namespace std;
 
void sel_sort(int p[],int size){
    for(int i=0;i<size-1;i++){
        int lc=i;
        for(int j=i+1;j<size;j++){
            if(p[lc]<p[j]){
                lc=j;
            }    
        }
        swap(p[lc],p[i]);
    }    
}  
 
int main(){
    int n,k;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&k);
        int *t=new int [k];
        for(int j=0;j<k;j++){
            scanf("%d",&t[j]);
        }
        sel_sort(t,k);
        int *nt=new int [k];
        for(int j=0;j<k;j++){
            nt[j]=(j+1)*t[j];
        }    
        sel_sort(nt,k);
        cout<<nt[0]<<endl;
  
        delete []t;
        delete []nt;
        
    }    
   
    return 0;
}