C. Replace To Make Regular Bracket Sequence

题目连接：

Description

You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.

The following definition of a regular bracket sequence is well-known, so you can be familiar with it.

Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.

For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.

Determine the least number of replaces to make the string s RBS.

Input

The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.

Output

If it's impossible to get RBS from s print Impossible.

Otherwise print the least number of replaces needed to get RBS from s.

Sample Input

[<}){}

Sample Output

2

Hint

题意

给你一个只含有括号的字符串，你可以将一种类型的左括号改成另外一种类型，右括号改成另外一种右括号

问你最少修改多少次，才能使得这个字符串匹配，输出次数

题解：

用stack，每次将左括号压进stack里面，遇到右括号就判断一下就好了

非法就很简单，看看栈最后是否还有，看看右括号的时候，左括号的栈是否为空

代码

#include
    
     using namespace std;string s;stack
     
       S;int main(){    cin>>s;    int ans = 0;    for(int i=0;i
      
       ')        {            if(S.size()==0)return puts("Impossible");            if(S.top()=='<')                S.pop();            else            {                ans++;                S.pop();            }        }        else if(s[i]=='}')        {            if(S.size()==0)return puts("Impossible");             if(S.top()=='{')                S.pop();            else            {                ans++;                S.pop();            }        }        else S.push(s[i]);    }    if(S.size()!=0)return puts("Impossible");    cout<
       
        <