description: There are n heaps of stone. The i-th heap has h_i stones. You want to change the number of stones in the heap by performing the following process once: 
  
    * You go through the heaps from the 3-rd heap to the n-th heap, in this order. 
    * Let i be the number of the current heap. 
    * You can choose a number d (0 ≤ 3 ⋅ d ≤ h_i), move d stones from the i-th heap to the (i - 1)-th heap, and 2 ⋅ d stones from the i-th heap to the (i - 2)-th heap. 
    * So after that h_i is decreased by 3 ⋅ d, h_{i - 1} is increased by d, and h_{i - 2} is increased by 2 ⋅ d. 
    * You can choose different or same d for different operations. Some heaps may become empty, but they still count as heaps. 
  
  
  
  What is the maximum number of stones in the smallest heap after the process?
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 2⋅ 10^5). Description of the test cases follows.
  
  The first line of each test case contains a single integer n (3 ≤ n ≤ 2 ⋅ 10^5).
  
  The second lines of each test case contains n integers h_1, h_2, h_3, …, h_n (1 ≤ h_i ≤ 10^9).
  
  It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case, print the maximum number of stones that the smallest heap can contain.
  
  Example
  
  Input
  
  
  4
  4
  1 2 10 100
  4
  100 100 100 1
  5
  5 1 1 1 8
  6
  1 2 3 4 5 6
  
  
  Output
  
  
  7
  1
  1
  3
  
  Note
  
  In the first test case, the initial heap sizes are [1, 2, 10, 100]. We can move the stones as follows. 
  
    * move 3 stones and 6 from the 3-rd heap to the 2-nd and 1 heap respectively. The heap sizes will be [7, 5, 1, 100]; 
    * move 6 stones and 12 stones from the last heap to the 3-rd and 2-nd heap respectively. The heap sizes will be [7, 17, 7, 82]. 
  
  
  
  In the second test case, the last heap is 1, and we can not increase its size.
  
  In the third test case, it is better not to move any stones.
  
  In the last test case, the final achievable configuration of the heaps can be [3, 5, 3, 4, 3, 3].
solution: import java.io.*;
  import java.util.*;
  
  public class C_BalancedStoneHeap_1600 {
      static int[] orig, arr;
      public static void main(String[] args) {
          MyScanner sc = new MyScanner();
          out = new PrintWriter(new BufferedOutputStream(System.out));
          int t = sc.nextInt();
          while(t-->0) {
              int n = sc.nextInt();
              arr = new int[n];
              orig = new int[n];
              for(int i = 0; i < n; i++) {
                  arr[i] = sc.nextInt();
                  orig[i] = arr[i];
              }
              for(int i = 0; i < n/2; i++) {
                  int temp = arr[i];
                  arr[i] = arr[n-i-1];
                  arr[n-i-1] = temp;
                  temp = orig[i];
                  orig[i] = orig[n-i-1];
                  orig[n-i-1] = temp;
              }
              System.out.println(search(1, 1000000000));
          }
  
          out.close();
      }
      static boolean possible(int size) {
          int[] copy = new int[arr.length];
          for(int i = 0; i < arr.length; i++) {
              copy[i] = arr[i];
          }
          for(int i = 0; i < copy.length-2; i++) {
              if(copy[i] < size) return false;
              int d = Math.min(orig[i]/3, (copy[i]-size)/3);
              copy[i] -= 3*d;
              copy[i+1] += d;
              copy[i+2] += 2*d;
          }
          if(copy[copy.length-2] >= size && copy[copy.length-1] >= size) return true;
          return false;
      }
      static int search(int min, int max) {
          if(min == max) return min;
          if(min + 1 == max) {
              if(possible(max)) return max;
              return min;
          }
          int mid = (min+max)/2;
          if(possible(mid)) return search(mid, max);
          else return search(min, mid-1);
      }
  
      public static PrintWriter out;
  
      public static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public MyScanner() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine(){
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
  
      }
      static class Pair {
          public int x,y;
          public Pair(int x, int y) {
              this.x = x;
              this.y = y;
          }
      }
  }
  