Replace To Make Regular Bracket Sequence 括号配对

好久没写文章了|chutzpah.xyz 

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 <s1>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.

Example

Input

[<}){}

Output

2

Input

{()}[]

Output

0

Input

]]

Output

Impossible

代码如下

#include<stdio.h>
#include<stack>
using namespace std;
char s[]="<>{}[]()",g;
int F(char c){
for(int i=0;i<8;i++)
if(c==s[i])
return i;
}
stack <char> S;
int main()
{
bool flag=true;
int ans=0;
while((g=getchar())&&g!='\n'){
if(flag){
if(F(g)%2==0) S.push(g);
else{
if(!S.empty()) {
if(S.top()&&F(g)-F(S.top())!=1)
ans++;
S.pop();
}else{
S.push('0');//随便入栈一个元素
flag=false;
}
}
}
}
if(S.empty()) printf("%d\n",ans);
else puts("Impossible");
while(!S.empty()) S.pop();
return 0;
}