We give the following inductive definition of a “regular brackets” sequence:

• the empty sequence is a regular brackets sequence,
• if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
• if a and b are regular brackets sequences, then ab is a regular brackets sequence.
• no other sequence is a regular brackets sequence

For instance, all of the following character sequences are regular brackets sequences:

(), [], (()), ()[], ()[()]

while the following character sequences are not:

(, ], )(, ([)], ([(]

Given a brackets sequence of characters a₁a₂ … a_n, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i₁, i₂, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im is a regular brackets sequence.

Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

//============================================================================
// Name        : POJ.cpp
// Author      :
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int MAXN=110;
char str[MAXN];
int dp[MAXN][MAXN];
int solve(int i,int j)
{if(dp[i][j]!=-1)return dp[i][j];if(j<=i)return dp[i][j]=0;else if(j==i+1){if( (str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']') )return dp[i][j]=2;else return dp[i][j]=0;}dp[i][j]=solve(i+1,j);for(int k=i+1;k<=j;k++)if( (str[i]=='('&&str[k]==')')||(str[i]=='['&&str[k]==']') )dp[i][j]=max(dp[i][j],2+solve(i+1,k-1)+solve(k+1,j));return dp[i][j];
}int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);while(scanf("%s",str)==1){if(strcmp(str,"end")==0)break;memset(dp,-1,sizeof(dp));int n=strlen(str);printf("%d\n",solve(0,n-1));}return 0;
}