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 <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

#include<cstdio> #include<cstring> #include<stack> using namespace std; char s[1000010]; bool cmp(char a,char b) { if((a=='['&&b==']')||(a=='('&&b==')')||(a=='{'&&b=='}')||(a=='<'&&b=='>')) return true; else return false; } int main() { while(scanf("%s",s)!=EOF) { stack<char> sta; int len=strlen(s); int k=0; bool flag=true; for(int i=0;i<len;i++) { if(s[i]=='<'||s[i]=='['||s[i]=='{'||s[i]=='(') sta.push(s[i]); else if(s[i]=='>'||s[i]=='}'||s[i]==']'||s[i]==')') { if(sta.empty()) { flag=false; break; } else if(!cmp(sta.top(),s[i])) { k++; } sta.pop(); } } if(!sta.empty()) flag=false; if(flag) printf("%d\n",k); else printf("Impossible\n"); } return 0; }