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 字符是否可以刚好匹配,若可以匹配 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]);

若是两端可以匹配的，而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，这样把所有区间都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;
}