Given an array A of length N, your task is to find the element which repeats in A maximum number of times as well as the corresponding count. In case of ties, choose the smaller element first.
-----Input-----
First line of input contains an integer T, denoting the number of test cases. Then follows description of T cases. Each case begins with a single integer N, the length of A. Then follow N space separated integers in next line.  Assume that 1 <= T <= 100, 1 <= N <= 100 and for all i in [1..N] : 1 <= A[i] <= 10000
-----Output-----
For each test case, output two space separated integers V & C. V is the value which occurs maximum number of times and C is its count.
-----Example-----
Input:
2
5
1 2 3 2 5
6
1 2 2 1 1 2
Output:
2 2
1 3
Description:
In first case 2 occurs twice whereas all other elements occur only once. 
In second case, both 1 and 2 occur 3 times but 1 is smaller than 2.
n=int(input())
while(n):
    size=int(input())
    a=[]
    a=input().split()
    i=0
    #b=len(a)*[0]
    while(i<len(a)):
        a[i]=int(a[i])
        #print 'hello'
        i+=1
    j=max(a)
    freq=[]
    freq=(j+1)*[0]
    k=0
    while(k<len(a)):
        if(a[k]!=0):
            freq[a[k]-1]+=1
            k+=1
    count=max(freq)
    key=freq.index(count)+1
    print(key,count)
    n-=1
            
          
    

n=int(input())
while(n):
    size=int(input())
    a=[]
    a=input().split()
    i=0
    #b=len(a)*[0]
    while(i<len(a)):
        a[i]=int(a[i])
        #print 'hello'
        i+=1
    j=max(a)
    freq=[]
    freq=(j+1)*[0]
    k=0
    while(k<len(a)):
        if(a[k]!=0):
            freq[a[k]-1]+=1
            k+=1
    count=max(freq)
    key=freq.index(count)+1
    print(key,count)
    n-=1
            
          
    
