Problem: Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com). 

It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots. 

You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result. 

Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at. 

Input

The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.

Output

Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).

In the ASCII table the symbols go in this order: . @ ab...z

Examples

Input

vasyaatgmaildotcom


Output

vasya@gmail.com


Input

dotdotdotatdotdotat


Output

dot..@..at


Input

aatt


Output

a@t

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class CF{public static void main(String[] args){S s=new S();s.s();s.output();}}class S{
    void s() {
        sout(NS().replaceAll("(?!dot$|^dot)dot", ".").replaceAll("(.)at(.+)", "$1@$2"));
    }

    BufferedReader stdin=new BufferedReader(new InputStreamReader(System.in));StringTokenizer st=new StringTokenizer("");final long LINF=Long.MAX_VALUE;
    StringBuilder output=new StringBuilder();final int INF=Integer.MAX_VALUE;void sout(Object x){output.append(x.toString()).append('\n');}
    void inLong(long[]a,int n){for(int i=0;i<n;i++)a[i]=NL();}int min(int i1,int i2){return i1<i2?i1:i2;}double ND(){return Double.parseDouble(NS());}
    long min(long i1,long i2){return i1<i2?i1:i2;}int max(int i1,int i2){return i1>i2?i1:i2;}long max(long i1,long i2){return i1>i2?i1:i2;}
    String NS(){while(!st.hasMoreTokens())try{st=new StringTokenizer(stdin.readLine());}catch(IOException ignored){}return st.nextToken();}
    int NI(){return Integer.parseInt(NS());}long NL(){return Long.parseLong(NS());}String NLn()throws IOException{return stdin.readLine();}
    int abs(int x){return x<0?-x:x;}long abs(long x){return x<0?-x:x;}void sout(){output.append('\n');}void out(Object x){output.append(x.toString());}
    int mod(int x,int mod){return(x+mod)%mod;}void output(){System.out.print(output);}void inInt(int[]a,int n){for(int i=0;i<n;i++)a[i]=NI();}
}