You are given n integers a_1, a_2, …, a_n. Find the maximum value of max(a_l, a_{l + 1}, …, a_r) ⋅ min(a_l, a_{l + 1}, …, a_r) over all pairs (l, r) of integers for which 1 ≤ l < r ≤ n.

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5).

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).

It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.

Output

For each test case, print a single integer — the maximum possible value of the product from the statement.

Example

Input


4
3
2 4 3
4
3 2 3 1
2
69 69
6
719313 273225 402638 473783 804745 323328


Output


12
6
4761
381274500335

Note

Let f(l, r) = max(a_l, a_{l + 1}, …, a_r) ⋅ min(a_l, a_{l + 1}, …, a_r).

In the first test case, 

  * f(1, 2) = max(a_1, a_2) ⋅ min(a_1, a_2) = max(2, 4) ⋅ min(2, 4) = 4 ⋅ 2 = 8. 
  * f(1, 3) = max(a_1, a_2, a_3) ⋅ min(a_1, a_2, a_3) = max(2, 4, 3) ⋅ min(2, 4, 3) = 4 ⋅ 2 = 8. 
  * f(2, 3) = max(a_2, a_3) ⋅ min(a_2, a_3) = max(4, 3) ⋅ min(4, 3) = 4 ⋅ 3 = 12. 



So the maximum is f(2, 3) = 12.

In the second test case, the maximum is f(1, 2) = f(1, 3) = f(2, 3) = 6.
import sys,os,time
from math import log,sqrt,floor
from os import path
import math
'''Input handler'''
oj=True
if path.exists('inp.txt'):
    print('YES')
    sys.stdin=open('inp.txt','r')
    sys.stdout=open('out.txt','w')
    oj=False
'''Constants'''
mod=1000000007
INF=float('inf')


'''Input functions'''
def readint():
    return int(sys.stdin.readline())
def readlst():
    return list(map(int,sys.stdin.readline().strip().split()))
def readmul():
    return map(int,sys.stdin.readline().strip().split())
def is_power_two(x):
    return ((x^(x-1)&x)==x)
def lcm(a,b):
    return a*b//gcd(a, b)
'''
use for seive
maxN=80000+10000+6
primes=[True]*(maxN+1)
'''
def seive():
    primes[0],primes[1]=False,False
    for i in range(2,maxN+1):
        for j in range(i*i,maxN+1,i):
            if(primes[j]):
                primes[j]=False
def power_2(n):
    return 1<<n

def ceil(a,b):
    return int((a+b-1)/b)

def isPerfectSquare(x):
    if(x >= 0):
        sr = sqrt(x)
        return (sr*sr == x)
    return false



def omkar():
    n=readint()
    l=readlst()
    mx=max(l)
    ans=0
    ind=[]
    for i in range(0,n):
        if(i-1>=0):
            ans=max(ans, l[i]*l[i-1])
        if(i+1<n):
            ans=max(ans, l[i]*l[i+1])
    print(ans)
    return

def main():
    tc=readint()
    #tc=1
    start=time.time()
    while(tc):
        omkar()
        tc-=1
    if not oj:
        total=time.time()-start
        print(str(total*1000)+'ms')
main()
