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.*;
import java.io.*;

public class DIYWoodenLadder {

    public static void main(String[] args) throws Exception {
        FastReader in = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int tc = in.nextInt();
        for (int t = 0; t < tc; t++) {
            int n = in.nextInt();
            int max1 = -1, max2 = -1;
            int arr[] = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = in.nextInt();
                if (arr[i] > max1) {
                    max2 = max1;
                    max1 = arr[i];
                    continue;
                }
                if (arr[i] > max2) {
                    max2 = arr[i];
                }
            }
            out.println(Math.min(max2 - 1, n - 2));
        }//tc
        out.flush();
    }//psvm
    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

        FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        FastReader(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException x) {
                System.out.println(x);
            }
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }

        String nextLine() throws IOException {
            return br.readLine();
        }

        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        boolean hasNext() throws IOException {
            String s = br.readLine();
            if (s == null) {
                return false;
            }
            st = new StringTokenizer(s);
            return true;
        }
    }
}//class
