Problem: You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.

A string is called a regular bracket sequence (RBS) if it's of one of the following types: 

  * empty string; 
  * '(' + RBS + ')'; 
  * '[' + RBS + ']'; 
  * RBS + RBS. 



where plus is a concatenation of two strings.

In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.

What is the maximum number of moves you can perform?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 ⋅ 10^5.

Output

For each testcase print a single integer — the maximum number of moves you can perform on a given string s.

Example

Input


5
()
[]()
([)]
)]([
)[(]


Output


1
2
2
0
1

Note

In the first example you can just erase the whole string.

In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.

In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.

In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.

In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it.

Solution (in JAVA):

Solution:

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

public class Ed98C {
	public static void solve(IO io) {
		String s = io.getWord();

		int pOpen = 0;
		int sOpen = 0;
		int count = 0;

		for (int i = 0; i < s.length(); i++) {
			char c = s.charAt(i);
			switch (c) {
			case '(':
				pOpen++;
				break;
			case ')':
				if (pOpen > 0) {
					count++;
					pOpen--;
				}
				break;
			case '[':
				sOpen++;
				break;
			case ']':
				if (sOpen > 0) {
					count++;
					sOpen--;
				}

				break;
			}

		}
		System.out.println(count);

	}

	public static void main(String[] args) {
		IO io = new IO();
		int t = io.getInt();
		for (int testIndex = 0; testIndex < t; testIndex++) {
			solve(io);
		}
		io.close();
	}

	static class IO extends PrintWriter {

		public IO() {
			super(new BufferedOutputStream(System.out));
			r = new BufferedReader(new InputStreamReader(System.in));
		}

		public IO(InputStream i) {
			super(new BufferedOutputStream(System.out));
			r = new BufferedReader(new InputStreamReader(i));
		}

		public IO(InputStream i, OutputStream o) {
			super(new BufferedOutputStream(o));
			r = new BufferedReader(new InputStreamReader(i));
		}

		public boolean hasMoreTokens() {
			return peekToken() != null;
		}

		public int getInt() {
			return Integer.parseInt(nextToken());
		}

		public double getDouble() {
			return Double.parseDouble(nextToken());
		}

		public long getLong() {
			return Long.parseLong(nextToken());
		}

		public String getWord() {
			return nextToken();
		}

		private BufferedReader r;
		private String line;
		private StringTokenizer st;
		private String token;

		private String peekToken() {
			if (token == null)
				try {
					while (st == null || !st.hasMoreTokens()) {
						line = r.readLine();
						if (line == null)
							return null;
						st = new StringTokenizer(line);
					}
					token = st.nextToken();
				} catch (IOException e) {
				}
			return token;
		}

		private String nextToken() {
			String ans = peekToken();
			token = null;
			return ans;
		}
	}
}