G - Replace To Make Regular Bracket Sequence

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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

Input

{()}[]

Output

Input

]]

Output

Impossible

遍历s，遇到左括号则进栈，遇到右括号则取栈顶元素出来判断两个括号是否匹配，不匹配修改次数加一，在遍历的过程中记录res=左括号数量-右括号数量，如果在遍历过程中出现res<0的情况或者在遍历完后res!=0则说明左右括号不匹配输出Impossible，否则存在合理修改方案输出修改次数即可 
Code

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
#define maxn 1111111
char c[maxn],s[maxn];
int res,ans;
int main()
{
    while(~scanf("%s",c))
    {
        res=ans=0;
        int len=strlen(c);
        for(int i=0;i<len;i++)
        {
            if(c[i]=='{'||c[i]=='['||c[i]=='('||c[i]=='<')s[++res]=c[i];
            else 
            {
                if(c[i]!=s[res]+1&&c[i]!=s[res]+2)ans++;
                res--;
            }
            if(res<0)break;
        }
        if(res) printf("Impossible\n");
        else printf("%d\n",ans);
    }
    return 0;
}