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 Contest1623C
  {
      static class InputReader {
          BufferedReader reader;
          StringTokenizer tokenizer;
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
          String next() { // reads in the next string
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
          public int nextInt() { // reads in the next int
              return Integer.parseInt(next());
          }
          public long nextLong() { // reads in the next long
              return Long.parseLong(next());
          }
          public double nextDouble() { // reads in the next double
              return Double.parseDouble(next());
              }
          }
      static InputReader r = new InputReader(System.in);
      static PrintWriter pw = new PrintWriter(System.out);
      static long mod = 1000000007; static long[] h; static int n;
      public static void main(String[] args)
      {
      	int t = r.nextInt();
      	while (t > 0)
      	{
      		t--;
      		n = r.nextInt();
      		h = new long[n];
      		for (int i = 0; i < n; i ++)
      		{
      			h[i] = r.nextLong();
      		}
      		pw.println(lastTrue(1,1000000000));
      	}
          pw.close();
      }
      static boolean f(long x)
      {
      	long[] hi = new long[n];
      	for (int i = 0; i < n; i ++)
      	{
      		hi[i] = h[i];
      	}
      	for (int i = n - 1; i >= 2; i --)
      	{
      		long d = Math.max(0, Math.min((hi[i]-x)/3,h[i]/3));
      		
      		if (hi[i] < x)
      		{
      			return false;
      		}
      		hi[i-1] += d;
      		hi[i-2] += 2*d;
      	}
      	if (hi[0] >= x && hi[1] >= x)
      	{
      		return true;
      	}
      	return false;
      }
      public static int lastTrue(int lo, int hi) {
          lo--;
          while (lo < hi) {
              int mid = lo + (hi - lo + 1) / 2;
              if (f(mid)) {
                  lo = mid;
              } else {
                  hi = mid - 1;
              }
          }
          return lo;
      }
  }