Yelisey has an array a of n integers.

If a has length strictly greater than 1, then Yelisei can apply an operation called minimum extraction to it: 

  1. First, Yelisei finds the minimal number m in the array. If there are several identical minima, Yelisey can choose any of them. 
  2. Then the selected minimal element is removed from the array. After that, m is subtracted from each remaining element. 



Thus, after each operation, the length of the array is reduced by 1.

For example, if a = [1, 6, -4, -2, -4], then the minimum element in it is a_3 = -4, which means that after this operation the array will be equal to a=[1 {- (-4)}, 6 {- (-4)}, -2 {- (-4)}, -4 {- (-4)}] = [5, 10, 2, 0].

Since Yelisey likes big numbers, he wants the numbers in the array a to be as big as possible.

Formally speaking, he wants to make the minimum of the numbers in array a to be maximal possible (i.e. he want to maximize a minimum). To do this, Yelisey can apply the minimum extraction operation to the array as many times as he wants (possibly, zero). Note that the operation cannot be applied to an array of length 1.

Help him find what maximal value can the minimal element of the array have after applying several (possibly, zero) minimum extraction operations to the array.

Input

The first line contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The next 2t lines contain descriptions of the test cases.

In the description of each test case, the first line contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the original length of the array a. The second line of the description lists n space-separated integers a_i (-10^9 ≤ a_i ≤ 10^9) — elements of the array a.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

Print t lines, each of them containing the answer to the corresponding test case. The answer to the test case is a single integer — the maximal possible minimum in a, which can be obtained by several applications of the described operation to it.

Example

Input


8
1
10
2
0 0
3
-1 2 0
4
2 10 1 7
2
2 3
5
3 2 -4 -2 0
2
-1 1
1
-2


Output


10
0
2
5
2
2
2
-2

Note

In the first example test case, the original length of the array n = 1. Therefore minimum extraction cannot be applied to it. Thus, the array remains unchanged and the answer is a_1 = 10.

In the second set of input data, the array will always consist only of zeros.

In the third set, the array will be changing as follows: [\color{blue}{-1}, 2, 0] → [3, \color{blue}{1}] → [\color{blue}{2}]. The minimum elements are highlighted with \color{blue}{blue}. The maximal one is 2.

In the fourth set, the array will be modified as [2, 10, \color{blue}{1}, 7] → [\color{blue}{1}, 9, 6] → [8, \color{blue}{5}] → [\color{blue}{3}]. Similarly, the maximum of the minimum elements is 5.
import java.io.*;
import java.util.*;

public class C {
    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;
        }
    }

    static FastReader s = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    private static int[] rai(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextInt();
        }
        return arr;
    }

    private static int[][] rai(int n, int m) {
        int[][] arr = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] = s.nextInt();
            }
        }
        return arr;
    }

    private static long[] ral(int n) {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextLong();
        }
        return arr;
    }

    private static long[][] ral(int n, int m) {
        long[][] arr = new long[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] = s.nextLong();
            }
        }
        return arr;
    }

    private static int ri() {
        return s.nextInt();
    }

    private static long rl() {
        return s.nextLong();
    }

    private static String rs() {
        return s.next();
    }

    static long gcd(long a,long b)
    {
        if(b==0)
        {
            return a;
        }
        return gcd(b,a%b);
    }

    static int gcd(int a,int b)
    {
        if(b==0)
        {
            return a;
        }
        return gcd(b,a%b);
    }

    static boolean isPrime(int n) {
        //check if n is a multiple of 2
        if (n % 2 == 0) return false;
        //if not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    static long mpow(long base,long pow)
    {
        long res=1;
        while(pow>0)
        {
            if(pow%2==1)
            {
                res=(res*base)%MOD;
            }
            pow>>=1;
            base=(base*base)%MOD;
        }
        return res;
    }

    static TreeSet<Long> set= new TreeSet<>();
    static void sieveOfEratosthenes(int n)
    {
        // Create a boolean array
        // "prime[0..n]" and
        // initialize all entries
        // it as true. A value in
        // prime[i] will finally be
        // false if i is Not a
        // prime, else true.
        boolean[] prime = new boolean[n + 1];
        for (int i = 0; i <= n; i++)
            prime[i] = true;

        for (int p = 2; p * p <= n; p++)
        {
            // If prime[p] is not changed, then it is a
            // prime
            if (prime[p])
            {
                // Update all multiples of p
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }

        // Print all prime numbers
        for (int i = 2; i <= n; i++)
        {
            if (prime[i])
            {
                for(int pow = 1;pow<=10;pow++) {
                    long val = (long) Math.pow((long)i, pow);
                    set.add(val);
                    if(val>1000)
                    {
                        break;
                    }
                }
            }
        }
    }

    static long smallestDivisor(long n)
    {
        // if divisible by 2
        if (n % 2 == 0)
            return 2;

        // iterate from 3 to sqrt(n)
        for (long i = 3; i * i <= n; i += 2) {
            if (n % i == 0)
                return i;
        }

        return n;
    }


    static long power(long x, long y, int p)
    {

        // Initialize result
        long res = 1;

        // Update x if it is more than or
        // equal to p
        x = x % p;

        while (y > 0) {

            // If y is odd, multiply x
            // with result
            if (y % 2 == 1)
                res = (res * x) % p;

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }

        return res;
    }

    // Returns n^(-1) mod p
    static long modInverse(long n, int p)
    {
        return power(n, p - 2, p);
    }

    // Returns nCr % p using Fermat's
    // little theorem.
    static long nCrModPFermat(int n, int r)
    {

        if (n<r)
            return 0;
        // Base case
        if (r == 0)
            return 1;

        // Fill factorial array so that we
        // can find all factorial of r, n
        // and n-r


        return (fac[n] * modInverse(fac[r], MOD)
                % MOD * modInverse(fac[n - r], MOD)
                % MOD)
                % MOD;
    }
    static int MOD = (int)(1e9+7);
    static long[] fac;
    static void populateFact(int n)
    {
        fac = new long[n + 1];
        fac[0] = 1;

        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % MOD;
    }

    static long calculate(int[] arr, int n, int k, int i)
    {
        if(i==n)
        {
            long xor = arr[0], and=arr[0];
            for(int x=1;x<n;x++)
            {
                xor^=arr[x];
                and&=arr[x];
            }
            if(and>=xor)
            {
                return 1;
            }
            return 0;
        }
        long count = 0;
        for(int j=0;j<(1<<k);j++)
        {
            arr[i]=j;
            count+=calculate(arr,n,k,i+1);
        }
        return count;
    }
    static long calc(int n, int k)
    {
        int[] arr = new int[n];
        return calculate(arr,n,k,0);
    }

    static int countRem(char[] arr, int l, int r, int n, char c)
    {
        int i=l,j=r;
        int count = 0;
        while(i<=j)
        {
            if(arr[i] != arr[j])
            {
                if(arr[i]==c)
                {
                    count++;
                    i++;
                }
                else if(arr[j]==c)
                {
                    count++;
                    j--;
                }
                else
                {
                    return -1;
                }
            }
            else
            {
                i++;j--;
            }
        }
        return count;
    }

    public static void main(String[] args) {
        StringBuilder ans = new StringBuilder();
        int t = ri();
//        int t = 1;
        while (t-- > 0)
        {
            int n=ri();
            PriorityQueue<Long> pq = new PriorityQueue<>();
            for(int i=0;i<n;i++)
            {
                pq.add(rl());
            }

            long res = pq.remove();
            long val = res;
            while(pq.size()>0)
            {
                long curr = pq.remove();
                long mid = curr - val;
                res=Math.max(res,mid);
                val +=mid;
            }
            ans.append(res).append("\n");
        }
        out.print(ans.toString());
        out.flush();
    }

}
/*
4 3
2 1
1 3
3 4
8
3
1 2 3
2 3 1
3
3
2 1 2
2 1 3
3
 */