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.util.*;
  import java.io.*;
  import java.lang.*;
  import java.rmi.dgc.Lease;
  
  // Problem - Multiples of 3 onn codeforces MNedium difficulty level 
  // or check on youtube channel CodeNCode
  
  public class Problem {
     //  static int Mod = 1000000007;
     static int Mod=998244353;
      static long dp[][];
      static int g[][];
      static int vis[][];
      static int ans;
  
      public static void main(String[] args) {
          MyScanner scan = new MyScanner();
          PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  
           int t = scan.nextInt();
          // int t = 1;
          while (t-- > 0) {
            int n= scan.nextInt();
            int a[]= new int[n];
            int max= Integer.MIN_VALUE;
            for(int i=0;i<n;i++){
                a[i]= scan.nextInt();
                max= Math.max(max,a[i]);
            }
  
            int start=1;
            int end =max;
  
            int ans=end;
  
            while(start<=end){
                 int mid= start+ (end-start)/2;
                 int flag =1;
                 int temp[]= Arrays.copyOfRange(a,0,n);
                 for(int i=n-1;i>=2;i--){
                     if(temp[i]<mid){
                         flag=0;
                         break;
                     }
                     int can_donate= Math.min(a[i],temp[i]-mid);
                     int d= can_donate/3;
                     temp[i-1]+=d;
                     temp[i-2]+=2*d;
                 }
                 if(temp[1]<mid || temp[0]<mid) flag=0;
  
                 if(flag==0){
                     end=mid-1;
                 }else{
                     start= mid+1;
                     ans=mid;
                 }
            
              }
  
              out.println(ans);
            
            
  
  
          }
  
          out.close();
      }
  
  
  
  
      static ArrayList<Integer> digits(long n){
          ArrayList<Integer> list= new ArrayList<>();
          while(n>0){
              list.add((int)(n%10l));
              n=n/10l;
          }
          Collections.reverse(list);
          return list;
      }
  
      
      int upper_bound(long[] arr, int key) {
          int i=0, j=arr.length-1;
          if (arr[j]<=key) return j+1;
          if(arr[i]>key) return i;
          while (i<j){
              int mid= (i+j)/2;
              if(arr[mid]<=key){
                  i= mid+1;
              }else{
                  j=mid;
              }
          }
          return i;
      }
  
      static class Pair {
          int l;
          int r;
  
          Pair(int l, int r) {
              this.l = l;
              this.r = r;
          }
  
          @Override
          public int hashCode() {
              final int prime = 31;
              int result = 1;
              result = prime * result + l;
              result = prime * result + r;
              return result;
          }
  
          @Override
          public boolean equals(Object obj) {
              if (this == obj)
                  return true;
              if (obj == null)
                  return false;
              if (getClass() != obj.getClass())
                  return false;
              Pair other = (Pair) obj;
              if (l != other.l)
                  return false;
              if (r != other.r)
                  return false;
              return true;
          }
  
          @Override
          public String toString() {
              return "Pair [l=" + l + ", r=" + r + "]";
          }
  
      }
  
      public static void reverse(int a[]) {
          int n = a.length;
          for (int i = 0; i < n / 2; i++) {
              int temp = a[i];
              a[i] = a[n - i - 1];
              a[n - i - 1] = temp;
          }
      }
  
      public static void sort(int[] array) {
          ArrayList<Integer> copy = new ArrayList<>();
          for (Integer i : array)
              copy.add(i);
          Collections.sort(copy);
          for (int i = 0; i < array.length; i++)
              array[i] = copy.get(i);
      }
  
      public static void sort(long[] array) {
          ArrayList<Long> copy = new ArrayList<>();
          for (Long i : array)
              copy.add(i);
          Collections.sort(copy);
          for (int i = 0; i < array.length; i++)
              array[i] = copy.get(i);
      }
  
      public static long gcd(long a, long b) {
          if (b == 0)
              return a;
          return gcd(b, a % b); // gcd(a,b) = gcd(a-b,b) or gcd(a,b) = gcd(b,a%b) where a>b
      }
  
      public static int gcd(int a, int b) {
          if (b == 0)
              return a;
          return gcd(b, a % b); // gcd(a,b) = gcd(a-b,b) or gcd(a,b) = gcd(b,a%b) where a>b
      }
  
      public static long lcm(long a, long b) {
          return (a / gcd(a, b)) * b;
      }
  
      static long power(long x, long y) {
          long temp;
          if (y == 0) // a^b= (a^b/2)^2 if b is even
              return 1; // = (a*a^(b-1)) if b is odd
          temp = power(x, y / 2);
          if (y % 2 == 0)
              return temp * temp;
          else
              return x * temp * temp;
      }
  
      static long mod(long x) {
          return ((x % Mod + Mod) % Mod);
      }
  
      static long add(long a, long b) {
          return mod(mod(a) + mod(b));
      }
  
      static int mod(int x) {
          return ((x % Mod + Mod) % Mod);
      }
  
      static int add(int a, int b) {
          return mod(mod(a) + mod(b));
      }
  
      static int sub(int a, int b) {
          return mod(mod(a) - mod(b) + Mod);
      }
  
      static long mul(long a, long b) {
          return mod(mod(a) * mod(b));
      }
  
      static long pow(long x, long y, long p) {
          long res = 1; // Initialize result
          x = x % p; // Update x if it is more than or equal to p
  
          if (x == 0)
              return 0; // In case x is divisible by p;
  
          while (y > 0) {
              // If y is odd, multiply x with result
              if ((y & 1) != 0)
                  res = (res * x) % p;
  
              // y must be even now
              y = y >> 1; // y = y/2
              x = (x * x) % p;
          }
          return res;
      }
  
      static long modInverse(long a, long m) {
          // int g = gcd(a, m);
          // if (g != 1)
          // System.out.println("Inverse doesn't exist");
          // else {
          // // If a and m are relatively prime, then modulo
          // // inverse is a^(m-2) mode m
          // return power(a, m - 2, m);
          // }
          return pow(a, m - 2, m);
      }
  
  }
  
  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;
      }
  
  }
  