(CodeForces

(CodeForces - 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

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 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

题目大意：给出一个括号序列，可以替换若干个类型相同的括号（类型相同意味着只能是左括号换左括号），问至少需要换几次可以构成规则括号序列。

思路：直接用栈模拟就行。注意：<[>]不是规则序列他需要换两次变成[<>]才是规则序列。

#include<cstdio>#include<stack>#include<cstring>using namespace std;const int maxn=1000005;char s[maxn];int main(){    while(~scanf("%s",s))    {        stack<char> st;        int len=strlen(s);        bool flag=true;        int ans=0;        for(int i=0;i<len;i++)        {            if(s[i]=='('||s[i]=='['||s[i]=='<'||s[i]=='{') st.push(s[i]);            else            {                if(!st.empty())                {                    char tmp=st.top();                    st.pop();                    if(tmp=='('&&s[i]!=')') ans++;                    if(tmp=='['&&s[i]!=']') ans++;                    if(tmp=='<'&&s[i]!='>') ans++;                    if(tmp=='{'&&s[i]!='}') ans++;                }                else                {                    flag=false;                }            }        }        if(!flag||!st.empty()) printf("Impossible\n");        else printf("%d\n",ans);    }    return 0;}