Drazil likes heap very much. So he created a problem with heap:

There is a max heap with a height h implemented on the array. The details of this heap are the following:

This heap contains exactly 2^h - 1 distinct positive non-zero integers. All integers are distinct. These numbers are stored in the array a indexed from 1 to 2^h-1. For any 1 < i < 2^h, a[i] < a[\left ⌊{i/2}\right ⌋].

Now we want to reduce the height of this heap such that the height becomes g with exactly 2^g-1 numbers in heap. To reduce the height, we should perform the following action 2^h-2^g times:

Choose an index i, which contains an element and call the following function f in index i:

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Note that we suppose that if a[i]=0, then index i don't contain an element.

After all operations, the remaining 2^g-1 element must be located in indices from 1 to 2^g-1. Now Drazil wonders what's the minimum possible sum of the remaining 2^g-1 elements. Please find this sum and find a sequence of the function calls to achieve this value.

Input

The first line of the input contains an integer t (1 ≤ t ≤ 70 000): the number of test cases.

Each test case contain two lines. The first line contains two integers h and g (1 ≤ g < h ≤ 20). The second line contains n = 2^h-1 distinct positive integers a[1], a[2], …, a[n] (1 ≤ a[i] < 2^{20}). For all i from 2 to 2^h - 1, a[i] < a[\left ⌊{i/2}\right ⌋].

The total sum of n is less than 2^{20}.

Output

For each test case, print two lines.

The first line should contain one integer denoting the minimum sum after reducing the height of heap to g. The second line should contain 2^h - 2^g integers v_1, v_2, …, v_{2^h-2^g}. In i-th operation f(v_i) should be called.

Example

Input


2
3 2
7 6 3 5 4 2 1
3 2
7 6 5 4 3 2 1


Output


10
3 2 3 1
8
2 1 3 1

Source: codeforces | Difficulty: VERY_HARD
Tags: Greedy algorithms

Data structures

Implementation

Constructive algorithms


Solutions:
import io
import os
import sys
fileno = sys.stdin.fileno()
input = io.BytesIO(os.read(fileno, os.fstat(fileno).st_size)).readline
T = int(input())

def trim(A, h, g):
	N = len(A)
	ans = []
	initial_root = 0
	items_to_leave = pow(2, g) - 1
	expected_ans_size = pow(2, h) - pow(2, g)
	while len(ans) < expected_ans_size:
		root = initial_root
		operations = []
		ok = False
		while True:
			lchi = root * 2 + 1
			rchi = lchi + 1
			lchv = A[lchi] if lchi < N else 0
			rchv = A[rchi] if rchi < N else 0
			if lchv == 0 and rchv == 0:
				if root >= items_to_leave:
					ok = True
					A[root] = 0
				break
			if lchv > rchv:
				operations.append((root, lchv))
				root = lchi
			else:
				operations.append((root, rchv))
				root = rchi
		if ok:
			for (index, value) in operations:
				A[index] = value
			ans.append(initial_root + 1)
		else:
			initial_root += 1
	S = 0
	for i in range(items_to_leave):
		S += A[i]
	print(S)
	print(' '.join(map(str, ans)))
for t in range(T):
	(h, g) = map(int, input().split())
	A = list(map(int, input().split()))
	trim(A, h, g)


import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
T = int(input())
for _ in range(T):
	(H, G) = map(int, input().split())
	A = [0] + list(map(int, input().split()))
	N = len(A)
	target_N = 2 ** G - 1
	target_ans_len = 2 ** H - 2 ** G
	Ans = []
	Roots = [True] * (N + 1)
	idx_Roots = 1
	while True:
		idx = idx_Roots
		st = []
		while True:
			idx_l = idx << 1
			idx_r = idx_l + 1
			st.append((idx, A[idx]))
			if idx_l >= N or A[idx_l] == A[idx_r] == 0:
				A[idx] = 0
				break
			elif A[idx_l] > A[idx_r]:
				A[idx] = A[idx_l]
				idx = idx_l
			else:
				A[idx] = A[idx_r]
				idx = idx_r
		if st[-1][0] <= target_N:
			for (idx, a) in st:
				A[idx] = a
				Roots[idx] = False
			while not Roots[idx_Roots]:
				idx_Roots += 1
		else:
			Ans.append(idx_Roots)
			if len(Ans) == target_ans_len:
				break
	print(sum(A))
	print(' '.join(map(str, Ans)))