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 <iostream>

#include <cstdlib>

#include <algorithm>

#include <ctime>

#include <cmath>

#include <string>

#include <cstring>

#include <stack>

#include <queue>

#include <list>

#include <vector>

#include <map>

#include <set>

using namespace std;

const int INF=0x3f3f3f3f;

const double eps=1e-;

const double PI=acos(-1.0);

#define maxn 1100

char a[maxn];

int dp[maxn][maxn];

int is(char b, char c)

{

    if(b == '(' && c == ')' || b == '[' && c == ']')

        return ;

    else

        return ;

}

int dfs(int st, int ed)

{

    //

    if(st > ed)

        return ;

    if(st == ed)

       return ;

    if(dp[st][ed] != -) return dp[st][ed];

    int res = dfs(st+, ed);

    for(int k = st+; k <= ed; k++)

        if(is(a[st],a[k]))

        {

            res = max(res,dfs(st+,k-) +  + dfs(k+,ed));

            flag = ;

        }

        dp[st][ed] = res;

    return dp[st][ed];

}

int main()

{

    while(~scanf("%s", a))

    {

        if(strcmp(a, "end") == )

            break;

        memset(dp, -, sizeof dp);

        int ed = strlen(a)-;

        printf("%d\n", dfs(, ed));

    }

    return ;

}