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;
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int inf = (1 << 30) - 1;
const ll INF = (1LL << 60) - 1;
const double pi = acos(-1.0);
const double EPS = 1e-10;
template <typename T>
bool chmax(T &x, const T &y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T &x, const T &y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
void in() {}
template <typename Head, typename... Tail>
void in(Head &&head, Tail &&...tail) {
  cin >> head;
  in(forward<Tail>(tail)...);
}
void out() { cout << '\n'; }
template <typename Head, typename... Tail>
void out(Head &&head, Tail &&...tail) {
  cout << head << ' ';
  out(forward<Tail>(tail)...);
}
void outn() {}
template <typename Head, typename... Tail>
void outn(Head &&head, Tail &&...tail) {
  cout << head << '\n';
  outn(forward<Tail>(tail)...);
}
template <typename T, typename U>
void in(pair<T, U> &p) {
  cin >> p.first >> p.second;
}
template <typename T, typename U>
void out(pair<T, U> p) {
  cout << p.first << ' ' << p.second << '\n';
}
using vi = vector<int>;
using vll = vector<ll>;
template <typename T>
void vin(vector<T> &a) {
  for (int i = 0; i < (int)a.size(); i++) cin >> a[i];
}
template <typename T>
void vout(const vector<T> &a) {
  for (auto &e : a) cout << e << ' ';
  cout << '\n';
}
template <typename T>
void voutn(const vector<T> &a) {
  for (auto &e : a) cout << e << '\n';
}
template <typename T>
void unique(vector<T> &a) {
  sort(a.begin(), a.end()), a.erase(unique(a.begin(), a.end()), a.end());
}
vector<int> iota(int n) {
  vector<int> ret(n);
  iota(ret.begin(), ret.end(), 0);
  return ret;
}
template <typename T>
vector<int> iota(const vector<T> &a, bool greater = false) {
  vector<int> ret = iota((int)a.size());
  sort(ret.begin(), ret.end(),
       [&](int i, int j) { return (a[i] < a[j]) ^ greater; });
  return ret;
}
struct io_setup {
  io_setup() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << fixed << setprecision(15);
  }
} io_setup;
const int MOD = 1000000007;
int h, g;
struct Graph {
  vector<vector<int>> es;
  vector<int> a;
  ll ans = 0;
  vector<int> keep;
  const int n;
  Graph(int n) : n(n) { es.resize(n), a.resize(n); }
  void add_edge(int from, int to, bool directed = false) {
    es[from].push_back(to);
    if (!directed) es[to].push_back(from);
  }
  void remove(int root, int d) {
    while (true) {
      if (a[root] == 0) return;
      int D = search(root, d);
      if (D <= g) break;
      keep.emplace_back(root);
      ans -= a[root], dfs(root);
    }
    remove(2 * root, d + 1), remove(2 * root + 1, d + 1);
  }
  int search(int root, int d) {
    int ptr = root;
    while (true) {
      if (ptr >= n / 2 || (a[2 * ptr] == 0 && a[2 * ptr + 1] == 0)) break;
      if (a[2 * ptr] > a[2 * ptr + 1])
        ptr = 2 * ptr, d++;
      else
        ptr = 2 * ptr + 1, d++;
    }
    return d;
  }
  void dfs(int ptr) {
    if (ptr >= n / 2 || (a[2 * ptr] == 0 && a[2 * ptr + 1] == 0))
      a[ptr] = 0;
    else if (a[2 * ptr] > a[2 * ptr + 1])
      a[ptr] = a[2 * ptr], dfs(2 * ptr);
    else
      a[ptr] = a[2 * ptr + 1], dfs(2 * ptr + 1);
  }
  void solve() {
    for (int i = 1; i <= n - 1; i++) ans += a[i];
    remove(1, 1);
    out(ans), vout(keep);
  }
};
int main() {
  int T;
  in(T);
  while (T--) {
    in(h, g);
    Graph G(1 << h);
    for (int i = 1; i <= (1 << h) - 1; i++) in(G.a[i]);
    G.solve();
  }
}