description: Mocha is a young girl from high school. She has learned so much interesting knowledge from her teachers, especially her math teacher. Recently, Mocha is learning about binary system and very interested in bitwise operation.
  
  This day, Mocha got a sequence a of length n. In each operation, she can select an arbitrary interval [l, r] and for all values i (0≤ i ≤ r-l), replace a_{l+i} with a_{l+i}  \&  a_{r-i} at the same time, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). This operation can be performed any number of times.
  
  For example, if n=5, the array is [a_1,a_2,a_3,a_4,a_5], and Mocha selects the interval [2,5], then the new array is [a_1,a_2 \&  a_5, a_3 \&  a_4, a_4 \&  a_3, a_5 \&  a_2].
  
  Now Mocha wants to minimize the maximum value in the sequence. As her best friend, can you help her to get the answer?
  
  Input
  
  Each test contains multiple test cases. 
  
  The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Each test case consists of two lines.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 100) — the length of the sequence.
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
  
  Output
  
  For each test case, print one integer — the minimal value of the maximum value in the sequence.
  
  Example
  
  Input
  
  
  4
  2
  1 2
  3
  1 1 3
  4
  3 11 3 7
  5
  11 7 15 3 7
  
  
  Output
  
  
  0
  1
  3
  3
  
  Note
  
  In the first test case, Mocha can choose the interval [1,2], then the sequence becomes [ 0, 0], where the first element is 1 \& 2, and the second element is 2 \& 1.
  
  In the second test case, Mocha can choose the interval [1,3], then the sequence becomes [ 1,1,1], where the first element is 1 \& 3, the second element is 1 \& 1, and the third element is 3 \& 1.
solution: 
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  
  
  import static javax.swing.UIManager.get;
  import static javax.swing.UIManager.getString;
  
  
  public class Main {
  
  
      static class Pair implements Comparable<Pair>{
          int x=0;
          int y=0;
          public Pair(int x1,int y1)
          { x=x1;
              y=y1;
          }
  
  
          @Override
          public int compareTo(Pair o) {
              return this.x-o.x;
          }
      }
  
      static boolean checkPallindrome(int n)
      {    ArrayList<Integer> list = new ArrayList<>();
          while(n>0)
          { list.add(n%10);
              n/=10;
          }
          int low=0,high=list.size()-1;
          while(low<=high)
          {     if(list.get(low)!=list.get(high))
              return false;
              low++;
              high--;
          }
          return true;
      }
  
  
      static boolean check(String s)
      {
  
          int low=0,high=s.length()-1;
          while(low<=high)
          {     if(s.charAt(low)!=s.charAt(high))
              return false;
              low++;
              high--;
          }
          return true;
      }
  
      class Node implements Comparable<Node>{
        int x=0,y=0,z=0;
          @Override
          public int compareTo(Node o) {
              return this.y-o.y;
          }
      }
  
      static  int min=Integer.MAX_VALUE;
      public static void main(String[] args) throws IOException {
          FastReader sc = new FastReader();
  //(1)very very important**(never take the first problem for granted, always check the test cases) take 5 minutes more and check the edge cases
          // 5 minutes will not decreases rating as much as a wrong submission does it is easy u just think with an open mind and u will surely get the answer
          //(2)let ur brain consume the problem don't just jump to the solution. after reading the problem take a pause 1 minute
          //(3)go through the example test cases and also at least two of ur own test cases.Think of testcases which are difficult(edge cases).dry run ur concept
  
          int t = sc.nextInt();
          while (t-- > 0) {
          int n=sc.nextInt();
          int[] a =sc.readArray(n);
          int min=a[0];
          for(int i=1;i<n;i++)
        min=min&a[i];
  System.out.println(min);
  
          }
  
  
  
  
      }
  
  
  
  
  
  // syntax of conditional operator y=(x==1)?1:0;
  //Things to check when u r getting wrong answer
  // 1-  check the flow of the code
  //2- If ur stuck read the problem once again
  //3- before submitting always check the output format of ur code
  //4- don't check standings until problem B is done
  //5- if u r thinking ur concept is correct but still u r getting wrong answer try to implement it in another way
  //6- By default, java interpret all numeral literals as 32-bit integer values.
  // If you want to explicitely specify that this is something bigger then 32-bit integer you should use suffix L for long values.    example   long a = 600851475143L
  
  //All the functions
  
  
  
      static void debug(String s)
      { System.out.println(s);
  
      }
  
      //collections.sort use merge sort instead of quick sort but arrays.sort use quicksort whose worst time complexity is O(n^2)
      static int[] sort(int[] a)
      {         ArrayList<Integer> list = new ArrayList<>();
          for(int i=0;i<a.length;i++)
              list.add(a[i]);
  
          Collections.sort(list);
          int ind=0;
          for(int x:list)
              a[ind++]=x;
          return a;
  
      }
  
      //function to print an array for debugging
      static void print(int[] a) {
          for (int i = 0; i < a.length; i++)
              System.out.print(a[i] + " ");
          System.out.println();
      }
  
      static void printc(char[] a) {
          for (int i = 0; i < a.length; i++)
              System.out.print(a[i] + " ");
          System.out.println();
      }
  
  
      //normal gcd function, always put the greater number as a and the smaller number as b
      static int gcd(int a, int b) {
          if (b == 0)
              return a;
          return gcd(b, a % b);
  
      }
  
      static int lcm(int a,int b)
      {
          return (a*b)/gcd(a,b);
      }
  
  
      //to find gcd and lcm for numbers of long data type
      static long gcdl(long a, long b) {
          if (b == 0)
              return a;
          return gcdl(b, a % b);
  
      }
  
      static long lcml(long a,long b)
      {
          return (a*b)/gcdl(a,b);
      }
  
  
      //Input Reader to read faster input
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              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;
          }
  
  
          int[] readArray(int n)
          {  int[] a = new int[n];
              for(int i=0;i<n;i++)
                  a[i]=nextInt();
  
              return a;
          }
      }
  
  
  
  }
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  