Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
auto start = std::chrono::system_clock::now();
inline void skj() {
  std::chrono::time_point<std::chrono::system_clock> end;
  end = std::chrono::system_clock::now();
  std::chrono::duration<double> elapsed_seconds = end - start;
  std::time_t end_time = std::chrono::system_clock::to_time_t(end);
  cerr << "Time taken " << elapsed_seconds.count() * 1000 << "\n";
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
inline long long binexp(long long a, long long b, long long m) {
  if (a == 0) {
    return 0;
  }
  long long res = 1;
  a %= m;
  while (b) {
    if (b & 1) res = (res * 1ll * a) % m;
    a = (a * 1ll * a) % m;
    b >>= 1;
  }
  return res;
}
inline long long binmul(long long a, long long b, long long m) {
  a %= m;
  long long res = 0;
  while (b) {
    if (b & 1) {
      res = (res + a) % m;
    }
    a = (a + a) % m;
    b >>= 1;
  }
  return res;
}
long long h, g;
inline void check(vector<long long> &arr, long long id,
                  multiset<pair<long long, long long> > &ms,
                  vector<bool> &isRem) {
  long long flg = 0;
  long long i = id;
  vector<long long> res;
  while (arr[id]) {
    ms.erase({arr[id], id});
    isRem[id] = 0;
    res.push_back(id);
    if (arr[2 * id] > arr[2 * id + 1]) {
      arr[id] = arr[2 * id];
      id = 2 * id;
    } else {
      arr[id] = arr[2 * id + 1];
      id = 2 * id + 1;
    }
  }
  reverse(res.begin(), res.end());
  for (long long i : res) {
    if (arr[i] == 0)
      isRem[i] = 0;
    else if (i > g) {
      isRem[i] = 1;
    } else {
      if (arr[2 * i] > arr[2 * i + 1])
        isRem[i] = isRem[2 * i];
      else
        isRem[i] = isRem[2 * i + 1];
    }
    if (isRem[i]) ms.insert({arr[i], i});
  }
}
long long M[1 << 21];
inline void solve_me_senpai() {
  cin >> h >> g;
  h = 1 << h;
  g = 1 << g;
  h--, g--;
  vector<long long> arr(2 * h + 5, 0);
  vector<long long> values;
  long long sum = 0;
  for (long long i = 1; i <= h; i++) {
    cin >> arr[i];
    values.push_back(arr[i]);
    M[arr[i]] = i;
    sum += arr[i];
  }
  sort(values.rbegin(), values.rend());
  vector<long long> res;
  for (long long i : values) {
    long long id = M[i];
    while (arr[2 * id] != 0 or arr[2 * id + 1] != 0) {
      if (arr[2 * id] > arr[2 * id + 1])
        id = 2 * id;
      else
        id = 2 * id + 1;
    }
    if (id > g) {
      sum -= i;
      res.push_back(M[i]);
    } else
      continue;
    id = M[i];
    while (arr[id]) {
      if (arr[2 * id] > arr[2 * id + 1]) {
        M[arr[2 * id]] = id;
        arr[id] = arr[2 * id];
        id = 2 * id;
      } else {
        M[arr[2 * id + 1]] = id;
        arr[id] = arr[2 * id + 1];
        id = 2 * id + 1;
      }
    }
  }
  cout << sum;
  cout << "\n";
  for (long long i : res) cout << i << " ";
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  long long a = 1;
  while (t--) {
    solve_me_senpai();
    a++;
  }
  return 0;
}