【原创】G - Replace To Make Regular Bracket Sequence【枚举（1.情况数）栈{应尽量将能合并的情况作合并处理}】

G - Replace To Make Regular Bracket Sequence

https://cn.vjudge.net/contest/173547#problem/G

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：

#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(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;