#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("-ffloat-store")
using namespace std;
const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403() {}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
class CMP {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    return !(a.first < b.first || (a.first == b.first && a.second <= b.second));
  }
};
void add(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt == m.end())
    m.insert({x, cnt});
  else
    jt->second += cnt;
}
void del(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt->second <= cnt)
    m.erase(jt);
  else
    jt->second -= cnt;
}
bool cmp(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
const long long int mod = 1000000007L;
long long int T, n, i, j, k, in, cnt, l, r, u, v, x, y;
long long int m, h, g;
string s;
vector<long long int> a, c;
bool canRemove(long long int pos) {
  if (a[2 * pos] == 0 && a[2 * pos + 1] == 0) return c[pos];
  if (a[2 * pos] > a[2 * pos + 1]) return canRemove(2 * pos);
  return canRemove(2 * pos + 1);
}
void remove(long long int pos) {
  if (a[2 * pos] == 0 && a[2 * pos + 1] == 0) {
    a[pos] = 0;
    while (pos >= 1) {
      c[pos]--;
      pos /= 2;
    }
    return;
  }
  if (a[2 * pos] > a[2 * pos + 1]) {
    a[pos] = a[2 * pos];
    return remove(2 * pos);
  }
  a[pos] = a[2 * pos + 1];
  return remove(2 * pos + 1);
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> h >> g;
    a.clear();
    a.resize(2 * (1 << h));
    c.clear();
    c.resize(2 * (1 << h), 0);
    a.push_back(0);
    long long int sum = 0;
    for (i = (1); i <= ((1 << h) - 1); ++i) {
      cin >> a[i];
      sum += a[i];
    }
    for (long long int i = (1 << h) - 1; i > 0; --i) {
      c[i] = (i >= (1 << g));
      c[i] += c[2 * i];
      c[i] += c[2 * i + 1];
    }
    long long int cnt = (1 << h) - (1 << g);
    vector<long long int> b;
    for (long long int i = 1; i < (1 << h); ++i) {
      if (cnt == 0) break;
      if (canRemove(i)) {
        b.push_back(i);
        sum -= a[i];
        remove(i);
        cnt--;
        i--;
      }
    };
    ;
    assert(cnt == 0);
    cout << sum << "\n";
    for (auto x : b) cout << x << " ";
    cout << "\n";
  }
  aryanc403();
  return 0;
}

### Prompt

Create a solution in CPP for the following 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:

<image>

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

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("-ffloat-store")
using namespace std;
const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403() {}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
class CMP {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    return !(a.first < b.first || (a.first == b.first && a.second <= b.second));
  }
};
void add(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt == m.end())
    m.insert({x, cnt});
  else
    jt->second += cnt;
}
void del(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt->second <= cnt)
    m.erase(jt);
  else
    jt->second -= cnt;
}
bool cmp(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
const long long int mod = 1000000007L;
long long int T, n, i, j, k, in, cnt, l, r, u, v, x, y;
long long int m, h, g;
string s;
vector<long long int> a, c;
bool canRemove(long long int pos) {
  if (a[2 * pos] == 0 && a[2 * pos + 1] == 0) return c[pos];
  if (a[2 * pos] > a[2 * pos + 1]) return canRemove(2 * pos);
  return canRemove(2 * pos + 1);
}
void remove(long long int pos) {
  if (a[2 * pos] == 0 && a[2 * pos + 1] == 0) {
    a[pos] = 0;
    while (pos >= 1) {
      c[pos]--;
      pos /= 2;
    }
    return;
  }
  if (a[2 * pos] > a[2 * pos + 1]) {
    a[pos] = a[2 * pos];
    return remove(2 * pos);
  }
  a[pos] = a[2 * pos + 1];
  return remove(2 * pos + 1);
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> h >> g;
    a.clear();
    a.resize(2 * (1 << h));
    c.clear();
    c.resize(2 * (1 << h), 0);
    a.push_back(0);
    long long int sum = 0;
    for (i = (1); i <= ((1 << h) - 1); ++i) {
      cin >> a[i];
      sum += a[i];
    }
    for (long long int i = (1 << h) - 1; i > 0; --i) {
      c[i] = (i >= (1 << g));
      c[i] += c[2 * i];
      c[i] += c[2 * i + 1];
    }
    long long int cnt = (1 << h) - (1 << g);
    vector<long long int> b;
    for (long long int i = 1; i < (1 << h); ++i) {
      if (cnt == 0) break;
      if (canRemove(i)) {
        b.push_back(i);
        sum -= a[i];
        remove(i);
        cnt--;
        i--;
      }
    };
    ;
    assert(cnt == 0);
    cout << sum << "\n";
    for (auto x : b) cout << x << " ";
    cout << "\n";
  }
  aryanc403();
  return 0;
}

```