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.
// Working program with FastReader 
import java.io.*;
import java.util.*;
public class hh
{ 
    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 final int MAXN = 100001;
    // stores smallest prime factor for every number
    static int spf[] = new int[MAXN];
    // Calculating SPF (Smallest Prime Factor) for every
    // number till MAXN.
    // Time Complexity : O(nloglogn)
    static void sieve()
    {
        spf[1] = 1;
        for (int i=2; i<MAXN; i++)

        // marking smallest prime factor for every
        // number to be itself.
            spf[i] = i;

        // separately marking spf for every even
        // number as 2
        for (int i=4; i<MAXN; i+=2)
            spf[i] = 2;

        for (int i=3; i*i<MAXN; i++)
        {
            // checking if i is prime
            if (spf[i] == i)
            {
                // marking SPF for all numbers divisible by i
                for (int j=i*i; j<MAXN; j+=i)

                // marking spf[j] if it is not
                // previously marked
                    if (spf[j]==j)
                        spf[j] = i;
            }
        }
    }
    // A O(log n) function returning primefactorization
    // by dividing by smallest prime factor at every step
    static int getFactorization(int x)
    {
        int c=0;
        while (x != 1)
        {
            c++;
            x = x / spf[x];
        }
        return c;
    }

    /*
    static int LowerBound(int a[], int x) { // x is the target value or key
    int l=-1,r=a.length;
    while(l+1<r) {
    int m=(l+r)>>>1;
    if(a[m]>=x) r=m;
    else l=m;
    }
    return r;
    }

    static int UpperBound(int a[], int x) {// x is the key or target value
    int l=-1,r=a.length;
    while(l+1<r) {
    int m=(l+r)>>>1;
    if(a[m]<=x) l=m;
    else r=m;
    }
    return l+1;
    }
     */
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }

    // method to return LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }

    public static int[] swap(int a[], int left, int right)
    {
        int temp = a[left];
        a[left] = a[right];
        a[right] = temp;
        return a;
    }

    public static int[] reverse(int a[], int left, int right)
    {
        // Reverse the sub-array
        while (left < right) {
            int temp = a[left];
            a[left++] = a[right];
            a[right--] = temp;
        }
        return a;
    }

    public static int[] findNextPermutation(int a[])
    {
        int last = a.length - 2;

        // find the longest non-increasing suffix
        // and find the pivot
        while (last >= 0) {
            if (a[last] < a[last + 1]) {
                break;
            }
            last--;
        }
        // If there is no increasing pair
        // there is no higher order permutation
        if (last < 0)
            return a;

        int nextGreater = a.length - 1;

        // Find the rightmost successor to the pivot
        for (int i = a.length - 1; i > last; i--) {
            if (a[i] > a[last]) {
                nextGreater = i;
                break;
            }
        }

        // Swap the successor and the pivot
        a = swap(a, nextGreater, last);

        // Reverse the suffix
        a = reverse(a, last + 1, a.length - 1);

        // Return true as the next_permutation is done
        return a;
    }

    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a)
            l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++)
            a[i] = l.get(i);
    }
    static void sort(long[] a) {
        ArrayList<Long> l = new ArrayList<>();
        for (long i : a)
            l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++)
            a[i] = l.get(i);
    }
    static double pow(double p,double tt)
    {
        double ii,q,r;
        q=1;
        r=p;
        while(tt>1)
        {
            for(ii=1;2*ii<=tt;ii*=2)
                p*=p;
            tt-=ii;
            q*=p;
            p=r;
        }
        if(tt==1)
            q*=r;
        return q;
    }

    static long pow(long p,long tt,long mod)
    {
        long ii,q,r;
        q=1l;
        r=p;
        while(tt>1)
        {
            for(ii=1l;2*ii<=tt;ii*=2l)
                p=((p%mod)*(p%mod))%mod;
            tt-=ii;
            q=((q%mod)*(p%mod))%mod;
            p=r;
        }
        if(tt==1)
            q=((q%mod)*(r%mod))%mod;
        return q;
    }

    static int factorial(int n)
    {
        return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
    }

    public static long primeFactors(long n)
    {
        long c=0l;
        long max=0l;
        long z=0l;
        //ArrayList <Integer> ll=new ArrayList<>();
        // Print the number of 2s that divide n
        while (n%2==0)
        {
            c++;
            n /= 2l;
            //ll.add(2);
        }
        if(c>max)
        {
            max=c;
            z=2;
        }
        // n must be odd at this point. So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            c=0;
            while (n%i == 0)
            {
                c++;
                n /= i;
                //ll.add(i);
            }
            if(c>max)
            {
                max=c;
                z=i;
            }
        }
        c=0;
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n > 2)
        {
            c++;
            //ll.add((int)n);
        }
        if(c>max)
        {
            max=c;
            z=n;
        }
        return z;
    }

    static void PrimeList(){
        int i,j;
        int sieve[]=new int[100001];
        for(i=2;i*i<=100000;i++)
        {
            if(sieve[i]==0)
            {
                for(j=i*i;j<=100000;j+=i)
                    sieve[j]=1;
            }
        }
        ArrayList<Integer> primes=new ArrayList<>();
        for(i=2;i<=100000;i++)
        {
            if(sieve[i]==0)
                primes.add(i);
        }
    }

    static int comp(int a[],int b[],int n)
    {
        int z=0;
        for(int i=0;i<n;i++)
        {
            if(a[i]!=b[i])
            {
                z=1;
                break;
            }
        }
        if(z==0)
            return 1;
        else
            return 0;
    }

    static boolean isPowerOfTwo(int x)
    {
        return x!=0 && ((x&(x-1))==0);
    }

    static long calc (int a[],int b[],int n,int mod)
    {
        long sum=0l;
        int i;
        HashMap<Long,Long>x=new HashMap<>();
        HashMap<Long,Long>y=new HashMap<>();
        for(i=0;i<n;i++)
        {
            x.put((long)a[i],(long)i);
            y.put((long)b[i],(long)i);
        }
        for(i=0;i<n;i++)
        {
            sum=(sum+Math.abs(x.get((long)i)-y.get((long)i))%mod)%mod;
        }
        return sum;
    }

    static int[] rotate(int arr[])
    {
        int x = arr[arr.length-1], i;
        for (i = arr.length-1; i > 0; i--)
            arr[i] = arr[i-1];
        arr[0] = x;
        return arr;
    }

    public static void main(String[] args) 
    {
        //try
        //{
        FastReader d=new FastReader();
        PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        StringBuilder sb=new StringBuilder();
        int t,i,j,k,l,r,n;
        int mod = (int) 1e9 + 7;

        int Inf=Integer.MAX_VALUE;
        int negInf=Integer.MIN_VALUE;
        t=d.nextInt();
        //t=1;
        String s,s1;
        //char ch1,ch2,ch3,ch4;
        long ans,c,z;

        while(t-->0)
        {
            z=c=0l;
            ans=0l;
            n=d.nextInt();
            long a[]=new long[n];
            for(i=0;i<n;i++)
                a[i]=d.nextLong();
            sort(a);
            long x=0l;
            if(n==1)
                pr.println(a[0]);
            else
            {
                for(i=0;i<n-1;i++)
                {
                    if((long)a[i+1]-a[i]-(long)x>=(long)a[i])
                    {
                        
                        a[i+1]=(long)a[i+1]-a[i]-(long)x;
                        x+=(long)a[i];
                    }
                    else
                    {
                        pr.println(a[i]);
                        z=1;
                        break;
                    }
                }
                if(z==0)
                pr.println(a[i]);
            }
        }
        /*
        }catch(Exception e) {
        System.out.println(0);
        }*/
        pr.flush();
    }
}