Let's denote a k-step ladder as the following structure: exactly k + 2 wooden planks, of which

  * two planks of length at least k+1 — the base of the ladder; 
  * k planks of length at least 1 — the steps of the ladder; 



Note that neither the base planks, nor the steps planks are required to be equal.

For example, ladders 1 and 3 are correct 2-step ladders and ladder 2 is a correct 1-step ladder. On the first picture the lengths of planks are [3, 3] for the base and [1] for the step. On the second picture lengths are [3, 3] for the base and [2] for the step. On the third picture lengths are [3, 4] for the base and [2, 3] for the steps. 

<image>

You have n planks. The length of the i-th planks is a_i. You don't have a saw, so you can't cut the planks you have. Though you have a hammer and nails, so you can assemble the improvised "ladder" from the planks.

The question is: what is the maximum number k such that you can choose some subset of the given planks and assemble a k-step ladder using them?

Input

The first line contains a single integer T (1 ≤ T ≤ 100) — the number of queries. The queries are independent.

Each query consists of two lines. The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of planks you have.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the lengths of the corresponding planks.

It's guaranteed that the total number of planks from all queries doesn't exceed 10^5.

Output

Print T integers — one per query. The i-th integer is the maximum number k, such that you can choose some subset of the planks given in the i-th query and assemble a k-step ladder using them.

Print 0 if you can't make even 1-step ladder from the given set of planks.

Example

Input


4
4
1 3 1 3
3
3 3 2
5
2 3 3 4 2
3
1 1 2


Output


2
1
2
0

Note

Examples for the queries 1-3 are shown at the image in the legend section.

The Russian meme to express the quality of the ladders:

<image>
import sys     
import math as mt
import bisect
input=sys.stdin.readline
t=int(input())
#t=1
def ncr_util():
    inv[0]=inv[1]=1
    fact[0]=fact[1]=1
    for i in range(2,300001):
        inv[i]=(inv[i%p]*(p-p//i))%p
    for i in range(1,300001):
        inv[i]=(inv[i-1]*inv[i])%p
        fact[i]=(fact[i-1]*i)%p
def solve():
    mini=min(l[0],l[1])
    n1=n-2
    return max(min(n1,mini-1),0)
    
    
for _ in range(t):
    n=int(input())
    #n1=int(input())
    #s=input()
    #n=int(input())
    #n,m=(map(int,input().split()))
    #n1=n
    #a=int(input())
    #b=int(input())
    #a,b,c,r=map(int,input().split())
    #x2,y2=map(int,input().split())
    #n=int(input())
    #s=input()
    #s1=input()
    #p=input()
                    
    l=list(map(int,input().split()))
    #l=str(n)
    l.sort(reverse=True)
    #l2.sort(reverse=True)
    #l1.sort(reverse=True)
    print(solve())            
        