CodeForces - 612C Replace To Make Regular Bracket Sequence 压栈

C. 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 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 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

该题意思大概就是要形成正确的括号形式，左括号之间可以相互变换，右括号之间也可以相互变换，但左右括号之间不能相互变换。求出最小变换次数。 可以用STL中的stack解决，但没学过就用数组进行模拟压栈。

#include<stdio.h>
#include<string.h> 
int main()
{
	char a[1000005],b[1000000];
	int n,j=0,sum=0,sum2=0;
	scanf("%s",a);
	n=strlen(a);
	for(int i=0;i<n;i++)
	{
		b[j]=a[i];
		if(sum2<0)
		{
		printf("Impossible\n");
		return 0;	
		}
		switch(a[i])//利用ASCII码进行判断，如果匹配则弹出
		{
			case '{':j++;sum2++;break;
			case '[':j++;sum2++;break;
			case '(':j++;sum2++;break;
			case '<':j++;sum2++;break;
			case '}':if(b[j-1]+2==a[i]){j--;}else{j--;sum++;}sum2--;break;
			case ']':if(b[j-1]+2==a[i]){j--;}else{j--;sum++;}sum2--;break;
			case ')':if(b[j-1]+1==a[i]){j--;}else{j--;sum++;}sum2--;break;
			case '>':if(b[j-1]+2==a[i]){j--;}else{j--;sum++;}sum2--;break; 
		}
	}
		if(sum2!=0)
		{
		printf("Impossible\n");
		return 0;	
		}
	printf("%d\n",sum);
	
	
	return 0;
}