Problem: The Balance of the World

The world should be finely balanced. Positive vs. negative, light vs. shadow, and left vs. right brackets. Your mission is to write a program that judges whether a string is balanced with respect to brackets so that we can observe the balance of the world.

A string that will be given to the program may have two kinds of brackets, round ("( )") and square ("[ ]"). A string is balanced if and only if the following conditions hold.

* For every left round bracket ("("), there is a corresponding right round bracket (")") in the following part of the string.
* For every left square bracket ("["), there is a corresponding right square bracket ("]") in the following part of the string.
* For every right bracket, there is a left bracket corresponding to it.
* Correspondences of brackets have to be one to one, that is, a single bracket never corresponds to two or more brackets.
* For every pair of corresponding left and right brackets, the substring between them is balanced.



Input

The input consists of one or more lines, each of which being a dataset. A dataset is a string that consists of English alphabets, space characters, and two kinds of brackets, round ("( )") and square ("[ ]"), terminated by a period. You can assume that every line has 100 characters or less. The line formed by a single period indicates the end of the input, which is not a dataset.

Output

For each dataset, output "yes" if the string is balanced, or "no" otherwise, in a line. There may not be any extra characters in the output.

Sample Input


So when I die (the [first] I will see in (heaven) is a score list).
[ first in ] ( first out ).
Half Moon tonight (At least it is better than no Moon at all].
A rope may form )( a trail in a maze.
Help( I[m being held prisoner in a fortune cookie factory)].
([ (([( [ ] ) ( ) (( ))] )) ]).
.
.


Output for the Sample Input


yes
yes
no
no
no
yes
yes






Example

Input




Output

Solution (in JAVA):

Solution:

import java.util.Scanner;
import java.util.Stack;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()) {
			String str = sc.nextLine();
			if(str.equals("."))break;
			Stack<String> b = new Stack<String>();
			boolean can = true;
			String ans = "yes";
			StringBuilder sb = new StringBuilder();
			for(int i=0;i<str.length();i++) {
				char a = str.charAt(i);
				if(a=='(' || a==')' || a=='[' || a==']') {
					sb.append(a);
				}
			}

			for(int i=0;i<sb.length();i++) {	
				String a = String.valueOf(sb.charAt(i));
				if(a.equals("(")) {
					if(b.empty() || b.peek().equals("(") || b.peek().equals("["))
						b.push(a);
					else {
						can = false;
						break;
					}
					
				}
				else if(a.equals(")")) {
					if(b.empty()) {
						can  = false;
						break;
					}
					else if(b.peek().equals("(") )
						b.pop();
					else{
						can = false;
						break;
					}
				}
				else if(a.equals("[")) {
					if(b.empty() || b.peek().equals("[") || b.peek().equals("("))
						b.push(a);
					else{
						can = false;
						break;
					}
				}
				else if(a.equals("]")) {
					if(b.empty()) {
						can = false;
						break;
					}
					else if(b.peek().equals("["))
						b.pop();
					else {
						can = false;
						break;
					}
				}
			}
			if(!can || !b.empty())ans="no";
			System.out.println(ans);
			b.clear();
		}
	}

}