Replace To Make Regular Bracket Sequence（括号匹配）

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 ands₂ 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 ofs does not exceed10⁶.

Output

If it's impossible to get RBS from s printImpossible.

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

Example

Input

[<}){}

Output

2

Input

{()}[]

Output

0

Input

]]

Output

Impossible

题意：符号配对，替换同等方向符号使其配对；

思路：刚开始看感觉很难，其实仔细想，就是简单的模拟 + 栈，代码有解析；

#include<stdio.h>
#include<string.h>
#include<stack>
using namespace std;
char a[1000000+10];
int main()
{
	int i,k;
	while(scanf("%s",a)!=EOF)
	{
		int len=strlen(a);
		if(len%2)//奇数不会最后成对 ； 
		{
			printf("Impossible\n");
			continue;
		}
		stack<char> s;
		bool flag=true;
		k=0;
		for(i=0;i<len;i++)
		{
			if(a[i]=='<' || a[i]=='(' || a[i]=='{' || a[i]=='[')//直接压入栈； 
				s.push(a[i]);
			else
			{
				if(a[i]=='>')
				{
					if(s.empty() || s.top()=='}' || s.top()==')' || s.top()==']') 
					{    //仔细思考，这四种情况可以终止循环了； 
						flag=false;
						break;
					}
					else if(s.top()=='<')
						s.pop();
					else if(s.top()=='{' || s.top()=='(' || s.top()=='[')
					{   //可以替换字符； 
						k++;
						s.pop();
					}
				}
				if(a[i]=='}')
				{
					if(s.empty() || s.top()=='>' || s.top()==')' || s.top()==']')
					{
						flag=false;
						break;
					}
					else if(s.top()=='{')
						s.pop();
					else if(s.top()=='<' || s.top()=='(' || s.top()=='[')
					{
						k++;
						s.pop();
					}
				}
				if(a[i]==']')
				{
					if(s.empty() || s.top()=='}' || s.top()==')' || s.top()=='>')
					{
						flag=false;
						break;
					}
					else if(s.top()=='[')
						s.pop();
					else if(s.top()=='{' || s.top()=='(' || s.top()=='<')
					{
						k++;
						s.pop();
					}
				}
				if(a[i]==')')
				{
					if(s.empty() || s.top()=='}' || s.top()=='>' || s.top()==']')
					{
						flag=false;
						break;
					}
					else if(s.top()=='(')
						s.pop();
					else if(s.top()=='{' || s.top()=='<' || s.top()=='[')
					{
						k++;
						s.pop();
					}
				}
			}
		}
		if(!s.empty() || !flag)
			printf("Impossible\n");
		else printf("%d\n",k);
	}
	return 0;
}