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

#
# inf = float("inf")
# # sys.setrecursionlimit(10000000)
# # abc='abcdefghijklmnopqrstuvwxyz'
# # abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
# # words = {1:'one',2:'two',3:'three',4:'four',5:'five',6:'six',7:'seven',8:'eight',9:'nine',10:'ten',11:'eleven',12:'twelve',13:'thirteen',14:'fourteen',15:'quarter',16:'sixteen',17:'seventeen',18:'eighteen',19:'nineteen',20:'twenty',21:'twenty one',22:'twenty two',23:'twenty three',24:'twenty four',25:'twenty five',26:'twenty six',27:'twenty seven',28:'twenty eight',29:'twenty nine',30:'half'}
# # vow=['a','e','i','o','u']
# # dx,dy=[0,1,0,-1],[1,0,-1,0]
#
# # import random
# from collections import deque, Counter, OrderedDict,defaultdict
# # from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
# # from math import ceil,floor,log,sqrt,factorial,pi,gcd,degrees,atan2
# # from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
# mod, MOD = 1000000007, 998244353
#
#
import queue

# def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
#
#
# def get_ints(): return map(int, sys.stdin.readline().strip().split())
#
#
# def input(): return sys.stdin.readline().strip()

for i in range(int(input())):
	n = int(input())
	a = list(map(int, input().split()))
	a.sort()
	k = min(a[-1], a[-2]) - 1
	while n < (k + 2):
		k = k - 1
	print(k)
