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.*;import java.lang.*;import java.util.*;
  //* --> number of prime numbers less then or equal to x are  -->  x/ln(x)
  //* --> String concatenation using the + operator within a loop should be avoided. Since the String object is immutable, each call for concatenation will
  // result in a new String object being created.
  // THE SIEVE USED HERE WILL RETURN A LIST CONTAINING ALL THE PRIME NUMBERS TILL N
   public class a {static FastScanner sc;static PrintWriter pw;static class FastScanner {InputStreamReader is;BufferedReader br;StringTokenizer st;
  public FastScanner() {is = new InputStreamReader(System.in);br = new BufferedReader(is);}
  String next() throws Exception {while (st == null || !st.hasMoreElements())st = new StringTokenizer(br.readLine());
  return st.nextToken();}int nextInt() throws Exception {return Integer.parseInt(next());}long nextLong() throws Exception {
  return Long.parseLong(next());}int[] readArray(int num) throws Exception {int arr[]=new int[num];
  for(int i=0;i<num;i++)arr[i]=nextInt();return arr;}String nextLine() throws Exception {return br.readLine();
  }} public static boolean power_of_two(int a){if((a&(a-1))==0){ return true;}return false;}
  static boolean PS(double x){if (x >= 0) {double i= Math.sqrt(x);if(i%1!=0){
  return false;}return ((i * i) == x);}return false;}public static int[] ia(int n){int ar[]=new int[n];
  return ar;}public static long[] la(int n){long ar[]=new long[n];return ar;}
  public static void print(int ans,int t){System.out.println("Case"+" "+"#"+t+":"+" "+ans);}
  static long mod=1000000007;static int max=Integer.MIN_VALUE;static int min=Integer.MAX_VALUE;
  public static void sort(long[] arr){//because Arrays.sort() uses quicksort which is dumb
  //Collections.sort() uses merge sort
  ArrayList<Long> ls = new ArrayList<Long>();for(long x: arr)ls.add(x);Collections.sort(ls);
  for(int i=0; i < arr.length; i++)arr[i] = ls.get(i);}public static long fciel(long a, long b) {if (a == 0)  return 0;return (a - 1) / b + 1;}
   static boolean[] is_prime = new boolean[1000001];static ArrayList<Integer> list = new ArrayList<>();
  static long n = 1000000;public static void sieve() {Arrays.fill(is_prime, true);
  is_prime[0] = is_prime[1] = false;for (int i = 2; i * i <= n; i++) {
  if (is_prime[i]) {for (int j = i * i; j <= n; j += i)is_prime[j] = false;}}for (int i = 2; i <= n; i++) {
  if (is_prime[i]) {list.add(i);}}}
  public static void main(String args[]) throws java.lang.Exception {
  sc = new FastScanner();pw = new PrintWriter(System.out);StringBuilder s = new StringBuilder();
  int t=sc.nextInt();
  while(t-->0)
  {
      int n=sc.nextInt();
      long ar[]=new long[n];
      for(int i=0;i<n;i++)
      {
          ar[i]=sc.nextLong();
      }
      sort(ar);
     long min=ar[0];
     for(int i=1;i<n;i++)
     {
  min=min&ar[i];
     }
  s.append(min);
  if(t>0)
  {
      s.append("\n");
  }
          }
  pw.print(s);pw.close();}}