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.io.*;
import java.util.*;

public class Main {
	
	static class Scanner
	{
	    BufferedReader buf;
	    StringTokenizer tok;
	    Scanner()
	    {
	        buf = new BufferedReader(new InputStreamReader(System.in));
	    }
	    
	    boolean hasNext()
	    {
	        while(tok == null || !tok.hasMoreElements()) 
	        {
	            try
	            {
	                tok = new StringTokenizer(buf.readLine());
	            } 
	            catch(Exception e) 
	            {
	                return false;
	            }
	        }
	        return true;
	    }
	    
	    String next()
	    {
	        if(hasNext()) return tok.nextToken();
	        return null;
	    }
	    
	    int nextInt()
	    {
	        return Integer.parseInt(next());
	    }
	    
	    long nextLong()
	    {
	        return Long.parseLong(next());
	    }
	    
	    double nextDouble()
	    {
	        return Double.parseDouble(next());
	    }
	    
	}

	
	public static void main(String[] args) {
		Scanner sc =new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = sc.nextInt();
		int[] ans = new int[t];
		while(--t >= 0) {
			int n = sc.nextInt();
			int[] arr = new int[n];
			int max1 = 1, max2 = 1;
			for(int i = 0; i < n; i++) {
				arr[i] = sc.nextInt();
				if(arr[i] > max1) {
					max2 = max1;
					max1 = arr[i];
				}
				else if(arr[i] > max2) {
					max2 = arr[i];
				}
			}
			int res = n-2;
			if(res <= 0)
				continue;
			res = Math.min(res,max2-1);
			ans[t] = res;
		}
		for(int i = ans.length-1; i >= 0; i--)
			System.out.println(ans[i]);
		out.close();
	}
	
    
   
}