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.

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<map>

#include<set>

#include<vector>

#include<cmath>

const int maxn=1e5+;

typedef long long ll;

using namespace std;

string str;

int dp[][];

int main()

{

    while(cin>>str)

    {

        if(str=="end")

        {

            break;

        }

        else

        {

            int n=str.length();

            for(int t=;t<n;t++)

            {

                dp[t][t]=;

            }

            for(int t=;t<n-;t++)

            {

                if((str[t]=='['&&str[t+]==']')||(str[t]=='('&&str[t+]==')'))

                {

                    dp[t][t+]=;

                }

                else

                {

                    dp[t][t+]=;

                }

            }

            for(int r=;r<=n;r++)

            {

                for(int i=;i<n;i++)

                {

                    int j=i+r-;

                    if(j>n)

                    break;

                    if((str[i]=='['&&str[j]==']')||(str[i]=='('&&str[j]==')'))

                    {

                        dp[i][j]=dp[i+][j-]+;

                    }

                    else

                    dp[i][j]=;

                    for(int k=i;k<j;k++)

                    {

                        dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+][j]);

                    }

                }

            }

            printf("%d\n",dp[][n-]);

        }

    }

    return ;

}