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。

#include
#include
#include
char s[1000010];
using namespace std;
int main()
{
	while(~scanf("%s",s))
	{
		int flag=0,sum=0,len=strlen(s);
		stack st;
		for(int i=0;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;
}