description: <image>
  
  William really likes puzzle kits. For one of his birthdays, his friends gifted him a complete undirected edge-weighted graph consisting of n vertices.
  
  He wants to build a spanning tree of this graph, such that for the first k vertices the following condition is satisfied: the degree of a vertex with index i does not exceed d_i. Vertices from k + 1 to n may have any degree.
  
  William wants you to find the minimum weight of a spanning tree that satisfies all the conditions.
  
  A spanning tree is a subset of edges of a graph that forms a tree on all n vertices of the graph. The weight of a spanning tree is defined as the sum of weights of all the edges included in a spanning tree.
  
  Input
  
  The first line of input contains two integers n, k (2 ≤ n ≤ 50, 1 ≤ k ≤ min(n - 1, 5)).
  
  The second line contains k integers d_1, d_2, …, d_k (1 ≤ d_i ≤ n). 
  
  The i-th of the next n - 1 lines contains n - i integers w_{i,i+1}, w_{i,i+2}, …, w_{i,n} (1 ≤ w_{i,j} ≤ 100): weights of edges (i,i+1),(i,i+2),…,(i,n).
  
  Output
  
  Print one integer: the minimum weight of a spanning tree under given degree constraints for the first k vertices. 
  
  Example
  
  Input
  
  
  10 5
  5 3 4 2 1
  29 49 33 12 55 15 32 62 37
  61 26 15 58 15 22 8 58
  37 16 9 39 20 14 58
  10 15 40 3 19 55
  53 13 37 44 52
  23 59 58 4
  69 80 29
  89 28
  48
  
  
  Output
  
  
  95
solution: #include <bits/stdc++.h>
  using namespace std;
  const int INFi = 1e9 + 5;
  const int md = 1e9 + 7;
  const long long INF = 2e18;
  const int maxN = 5e5 + 1;
  double getTime() { return clock() / (double)CLOCKS_PER_SEC; }
  inline int add(const int &a, const int &b) {
    return a + b >= md ? a + b - md : a + b;
  }
  inline int sub(const int &a, const int &b) {
    return a - b < 0 ? a - b + md : a - b;
  }
  inline int mul(const int &a, const int &b) { return (1ll * a * b) % md; }
  int binpow(int a, int b) {
    if (b <= 0) return 1;
    if (b % 2) return mul(a, binpow(a, b - 1));
    int m = binpow(a, b / 2);
    return mul(m, m);
  }
  int rev(int a) { return binpow(a, md - 2); }
  int n, k;
  vector<int> d;
  vector<vector<double>> g;
  vector<int> deg;
  double last = 0;
  void solve1() {
    last = 0;
    deg.assign(n, 0);
    vector<bool> was(n);
    vector<double> mn(n, INF);
    vector<int> fr(n, -1);
    mn[0] = 0;
    for (int _ = 0; _ < (n); ++_) {
      int x = -1;
      for (int j = 0; j < (n); ++j) {
        if (was[j]) continue;
        if (mn[j] < mn[x] || x == -1) {
          x = j;
          continue;
        }
        if (mn[j] == mn[x] && fr[j] >= k && j >= k) {
          x = j;
        }
      }
      if (fr[x] != -1) {
        deg[x]++;
        deg[fr[x]]++;
      }
      last += mn[x];
      was[x] = true;
      for (int j = 0; j < (n); ++j) {
        if (was[j]) continue;
        if (mn[j] > g[x][j]) {
          mn[j] = g[x][j];
          fr[j] = x;
        } else if (mn[j] == g[x][j] && x >= k) {
          fr[j] = x;
        }
      }
    }
  }
  int K = 10;
  bool need = false;
  void go(int i) {
    if (i == k) {
      solve1();
      return;
    }
    double l = 0, r = 110;
    for (int _ = 0; _ < (K); ++_) {
      double mid = (r + l) / 2;
      for (int j = 0; j < (n); ++j) {
        g[i][j] += mid;
        g[j][i] += mid;
      }
      go(i + 1);
      for (int j = 0; j < (n); ++j) {
        g[i][j] -= mid;
        g[j][i] -= mid;
      }
      if (deg[i] < d[i]) {
        r = mid;
      } else {
        l = mid;
      }
    }
    if (need) r = l;
    for (int j = 0; j < (n); ++j) {
      g[i][j] += r;
      g[j][i] += r;
    }
    go(i + 1);
    for (int j = 0; j < (n); ++j) {
      g[i][j] -= r;
      g[j][i] -= r;
    }
    last -= deg[i] * r;
  }
  void solve() {
    cin >> n >> k;
    d.resize(k);
    for (int i = 0; i < (k); ++i) cin >> d[i];
    g.resize(n, vector<double>(n));
    vector<int> ord(n);
    vector<long long> d2(n, n);
    for (int i = 0; i < (k); ++i) d2[i] = d[i];
    iota((ord).begin(), (ord).end(), 0ll);
    sort((ord).begin(), (ord).end(), [&](const int &i, const int &j) {
      return make_pair(d2[i], i) < make_pair(d2[j], j);
    });
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    for (int l = 0, r = 0; l < n; l = r) {
      while (r < n && d2[ord[r]] == d2[ord[l]]) r++;
      if (l < k) r = min(r, k);
    }
    vector<int> pos(n);
    for (int i = 0; i < (n); ++i) pos[ord[i]] = i;
    for (int i = 0; i < (n); ++i) {
      for (int j = i + 1; j < n; ++j) {
        cin >> g[pos[i]][pos[j]];
        g[pos[j]][pos[i]] = g[pos[i]][pos[j]];
      }
    }
    for (int i = 0; i < (k); ++i) d[i] = d2[ord[i]];
    need = true;
    go(0);
    for (int i = 0; i < (k); ++i) d[i]++;
    need = false;
    double L = last;
    go(0);
    double R = last;
    cout << round((R + L) / 2) << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int tests = 1;
    for (int _ = 0; _ < (tests); ++_) {
      solve();
    }
    return 0;
  }
  