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.BufferedReader;
  import java.io.File;
  import java.io.FileInputStream;
  import java.io.FileNotFoundException;
  import java.io.FileOutputStream;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.math.BigInteger;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.BitSet;
  import java.util.Calendar;
  import java.util.Collections;
  import java.util.Comparator;
  import java.util.HashMap;
  import java.util.HashSet;
  import java.util.LinkedList;
  import java.util.PriorityQueue;
  import java.util.SortedSet;
  import java.util.Stack;
  import java.util.StringTokenizer;
  import java.util.TreeMap;
  import java.util.TreeSet;
  
  /**
   * #
   *
   * @author pttrung
   */
  public class C_Round_763_Div2 {
      public static int MOD = 998244353;
  
      static double[][][] dp;
  
  
      public static void main(String[] args) throws FileNotFoundException {
          // PrintWriter out = new PrintWriter(new FileOutputStream(new File(
          // "output.txt")));
          PrintWriter out = new PrintWriter(System.out);
          Scanner in = new Scanner();
          int T = in.nextInt();
          for (int z = 0; z < T; z++) {
  
              int n = in.nextInt();
              long[] data = new long[n];
  
              for (int i = 0; i < n; i++) {
                  data[i] = in.nextInt();
              }
              long st = 0;
              long ed = Integer.MAX_VALUE;
              long re = 0;
              while (st <= ed) {
                  long mid = (st + ed) / 2;
                  boolean ok = true;
                  long[] cur = new long[n];
                  for (int i = n - 1; i >= 0; i--) {
                      if (cur[i] + data[i] < mid) {
                          ok = false;
                          break;
                      }
                      if (i < 2) {
                          continue;
                      }
                      long more = Long.min(data[i], cur[i] + data[i] - mid);
                      long tmp = more % 3;
                      more -= tmp;
                      cur[i - 1] += more / 3;
                      cur[i - 2] += 2 * more / 3;
                  }
  
                  if (ok) {
                      re = mid;
                      st = mid + 1;
                  } else {
                      ed = mid - 1;
                  }
              }
              out.println(re);
  
          }
  
          out.close();
      }
  
  
      static int abs(int a) {
          return a < 0 ? -a : a;
      }
  
  
      public static int[] KMP(String val) {
          int i = 0;
          int j = -1;
          int[] result = new int[val.length() + 1];
          result[0] = -1;
          while (i < val.length()) {
              while (j >= 0 && val.charAt(j) != val.charAt(i)) {
                  j = result[j];
              }
              j++;
              i++;
              result[i] = j;
          }
          return result;
  
      }
  
      public static boolean nextPer(int[] data) {
          int i = data.length - 1;
          while (i > 0 && data[i] < data[i - 1]) {
              i--;
          }
          if (i == 0) {
              return false;
          }
          int j = data.length - 1;
          while (data[j] < data[i - 1]) {
              j--;
          }
          int temp = data[i - 1];
          data[i - 1] = data[j];
          data[j] = temp;
          Arrays.sort(data, i, data.length);
          return true;
      }
  
      public static int digit(long n) {
          int result = 0;
          while (n > 0) {
              n /= 10;
              result++;
          }
          return result;
      }
  
      public static double dist(long a, long b, long x, long y) {
          double val = (b - a) * (b - a) + (x - y) * (x - y);
          val = Math.sqrt(val);
          double other = x * x + a * a;
          other = Math.sqrt(other);
          return val + other;
  
      }
  
      public static class Point {
  
          int x;
          int y;
  
          public Point(int start, int end) {
              this.x = start;
              this.y = end;
          }
  
          public String toString() {
              return x + " " + y;
          }
  
      }
  
      public static class FT {
  
          long[] data;
  
          FT(int n) {
              data = new long[n];
          }
  
          public void update(int index, long value) {
              while (index < data.length) {
                  data[index] += value;
                  index += (index & (-index));
              }
          }
  
          public long get(int index) {
  
              long result = 0;
              while (index > 0) {
                  result += data[index];
                  index -= (index & (-index));
              }
              return result;
  
          }
      }
  
      public static long gcd(long a, long b) {
          if (b == 0) {
              return a;
          }
          return gcd(b, a % b);
      }
  
      public static long pow(long a, int b) {
          if (b == 0) {
              return 1;
          }
          if (b == 1) {
              return a;
          }
  
          long val = pow(a, b / 2);
          if (b % 2 == 0) {
              return (val * val);
          } else {
              return (val * ((val * a)));
  
          }
      }
  
      static class Scanner {
  
          BufferedReader br;
          StringTokenizer st;
  
          public Scanner() throws FileNotFoundException {
              // System.setOut(new PrintStream(new BufferedOutputStream(System.out), true));
              br = new BufferedReader(new InputStreamReader(System.in));
              //br = new BufferedReader(new InputStreamReader(new FileInputStream(new File("input.txt"))));
          }
  
          public String next() {
  
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (Exception e) {
                      throw new RuntimeException();
                  }
              }
              return st.nextToken();
          }
  
          public long nextLong() {
              return Long.parseLong(next());
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public double nextDouble() {
              return Double.parseDouble(next());
          }
  
          public String nextLine() {
              st = null;
              try {
                  return br.readLine();
              } catch (Exception e) {
                  throw new RuntimeException();
              }
          }
  
          public boolean endLine() {
              try {
                  String next = br.readLine();
                  while (next != null && next.trim().isEmpty()) {
                      next = br.readLine();
                  }
                  if (next == null) {
                      return true;
                  }
                  st = new StringTokenizer(next);
                  return st.hasMoreTokens();
              } catch (Exception e) {
                  throw new RuntimeException();
              }
          }
      }
  }
  