Let us define a regular brackets sequence in the following way:



1. Empty sequence is a regular sequence.



2. If S is a regular sequence, then (S) and [S] are both regular sequences.



3. If A and B are regular sequences, then AB is a regular sequence.



For example, all of the following sequences of characters are regular brackets sequences:



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



And all of the following character sequences are not:



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



Some sequence of characters '(', ')', '[', and ']' is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 ... an is called a subsequence of the string b1 b2
... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n.

Input



The input file contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.

Output



Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank
line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

1



([(]

Sample Output



()[()]

关键在于输入与输出格式。神坑。

区间dp,dp[i][j]表示
区间 i 到j之间的匹配数，区间两端的 字符能否够刚好匹配,若能够匹配 状态转移就多了一个 dp[i][j] = max(dp[i][k]+dp[k+1][j],dp[i+1][j-1]+1),若不能匹配就是dp[i][j]
= max(dp[i][j],dp[i][k]+dp[k+1][j]);

若是两端能够匹配的，并且两端匹配了导致的dp值最大那么就标记一下。mark[i][j]
= -1,否则 就mark[i][j] = k，这样把全部区间都dp一遍，回头再用DFS寻找。若是两端匹配导致值最大的 那么就直接输出这个字符标记一下，继续往更小的区间去搜索，否则 就分开两个区间搜索 [i,k]        [k+1,j]

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;

#define up(i,x,y) for(i=x;i<=y;i++)

#define down(i,x,y) for(i=x;i>=y;i--)

#define mem(a,b) memset(a,b,sizeof(a))

#define w(a) while(a)

char str[105];

int t,len,dp[105][105],mark[105][105],pos[105];

void dfs(int i,int j)

{

    if(mark[i][j]==-1)

    {

        pos[i]=pos[j]=1;

        dfs(i+1,j-1);

    }

    else if(mark[i][j]>=0)

    {

        dfs(i,mark[i][j]);

        dfs(mark[i][j]+1,j);

    }

    return;

}

int main()

{

    int l,i,j,k;

    scanf("%d%*c%*c",&t);

    while(t--)

    {

        gets(str);

        len=strlen(str);

        if(!len)

        {

            printf("\n");

            if(t)

            printf("\n");

            continue;

        }

        up(i,0,len-1)

        up(j,0,len-1)

        {

            mark[i][j]=-2;

            dp[i][j]=0;

        }

        mem(pos,0);

        i=j=l=0;

        w(l<len)

        {

            if(i==j)

            {

                i++,j++;

                if(j==len)

                    i=0,l++,j=l;

                continue;

            }

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

            {

                up(k,i,j-1)

                {

                    if(dp[i][j]<dp[i][k]+dp[k+1][j])

                    {

                        mark[i][j]=k;

                        dp[i][j]=dp[i][k]+dp[k+1][j];

                    }

                }

                if(dp[i][j]<dp[i+1][j-1]+1)

                {

                    mark[i][j]=-1;

                    dp[i][j]=dp[i+1][j-1]+1;

                }

            }

            else

            {

                up(k,i,j-1)

                {

                    if(dp[i][j]<dp[i][k]+dp[k+1][j])

                    {

                        mark[i][j]=k;

                        dp[i][j]=dp[i][k]+dp[k+1][j];

                    }

                }

            }

            i++,j++;

            if(j==len)

            {

                l++;

                i=0;

                j=l;

            }

        }

        dfs(0,len-1);

        up(i,0,len-1)

        {

            if(pos[i]==1)

                printf("%c",str[i]);

            else if(str[i]=='('||str[i]==')')

                printf("()");

            else

                printf("[]");

        }

        printf("\n");

        if(t)

        {

            printf("\n");

            getchar();

        }

    }

    return 0;

}