Codeforce 612C

 Replace To Make Regular Bracket Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Problem 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 <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.

Examples

input

[<}){}

output

2

input

{()}[]

output

0

input

]]

output

Impossible

题解：运用栈的特点就好了。

<span style="font-family:SimSun;font-size:18px;">#include<stdio.h>#include<string.h>#include<stack>char s[1000010];using namespace std;int main(){while(~scanf("%s",s)){int flag=0,sum=0,len=strlen(s);stack<char> st;for(int i=0;i<len;i++){if(s[i]=='('||s[i]=='{'||s[i]=='['||s[i]=='<') //左边的括号都进栈 {st.push(s[i]);}else if(!st.empty())  {if(s[i]==')'&&st.top()!='('||s[i]=='}'&&st.top()!='{'||s[i]==']'&&st.top()!='['||s[i]=='>'&&st.top()!='<'){sum++;   //若为右边的括号并且与它前一个的括号不配对则需要改动一次 }st.pop();    //这时栈顶的括号一定是能配对的，so出栈 }else{printf("Impossible\n"); //说明右边的括号多余了 flag=1;break;}}if(!flag&&st.empty()){printf("%d\n",sum);}if(!st.empty())   //判断栈是否为空，如果不为空说明左边的括号多余 printf("Impossible\n");}return 0;}</span>