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 java.util.Arrays;
import java.util.List;
import java.util.Scanner;
 
public class Watermelon {
 
    public static void main(String[] args) {
 
        Scanner sc = new Scanner(System.in);
        
        int q =sc.nextInt();
        
       for(int i=0;i<q;i++)
       {
           
           int n= sc.nextInt();
           int a[]= new int [n];
           int max1=Integer.MIN_VALUE , max2=Integer.MIN_VALUE , max1j=0 , max2j=0 , t=0;
           for (int j = 0; j < a.length; j++) {
               a[j]=sc.nextInt();
               
                if(a[j] > max1)
                {
                    max1=a[j];
                    
                    max1j=j;
                }
              
                
           }
           
           for (int j = 0; j < a.length; j++) {
              
               
                if(a[j] >= max2 && j!=max1j)
                {
                    max2=a[j];
                    
                    max2j=j;
                }
              
                
           }
           
           
           
           
           
           int k=n-2;
           
           while(k>0)
           {
               if(max1>=k+1 && max2>=k+1)
               {
                   break;   
               }
              
               
               k=k-1;
               
           }
           System.out.println(k);
           
           
           
           
           
       }
        
       
           
        
         
        
        
          
    
    
    
    
    
    }
    
}