1430_B. Barrels
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.

You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them. 

Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.

Some examples: 

  * if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10; 
  * if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0. 

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5) — the number of barrels and the number of pourings you can make.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{9}), where a_i is the initial amount of water the i-th barrel has.

It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.

Example

Input


2
4 1
5 5 5 5
3 2
0 0 0


Output


10
0
```python
def solve():
    n,k=map(int,input().split())
    if n==1:
        print(0)
        return
    ls=list(map(int,input().split()))
    ls.sort()
    res=ls.pop()
    for i in range(k):
        if ls:
            res+=ls.pop()
    print(res)
    return 
    
    
for _ in range(int(input())):
    solve()
```