A bracket sequence is a string, containing only characters "(", ")", "[" and "]".

A correct bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()[]", "([])" are correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.

A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.

You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.

Input

The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.

Output

In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.

Examples

Input

([])


Output

1
([])


Input

(((


Output

0
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.StringTokenizer;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskA solver = new TaskA();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskA {
    private String s;
    private int[] cnt, stack;
    private int n, top, res, resl, resr;

    private void Check(int l, int r) {
        if (res < cnt[r] - cnt[l]) {
            res = cnt[r] - cnt[l];
            resr = r; resl = l;
        }
    }

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        s = in.nextLine();
        s = " " + s;

        n = s.length() - 1;
        top = 0;
        cnt = new int[n + 1];
        stack = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            cnt[i] = cnt[i - 1] + (s.charAt(i) == '[' ? 1 : 0);
            if (s.charAt(i) == '(' || s.charAt(i) == '[') stack[++top] = i;
            else if ((s.charAt(i) == ']' && s.charAt(stack[top]) != '[') ||
                    (s.charAt(i) == ')' && s.charAt(stack[top]) != '(')) {
                top = 0;
                stack[++top] = i;
            } else {
                --top;
                Check(stack[top], i);
            }
        }

        out.println(res);
        if (res > 0)
            for(int i = resl + 1; i <= resr; ++i) out.print(s.charAt(i));
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        tokenizer = null;
    }

    public String nextLine() {
        try {
            String line = reader.readLine();
            if (line == null)
                throw new UnknownError();
            return line;
        } catch (Exception e) {
            throw new UnknownError(e.getMessage());
        }
    }

}

