hiho#1082 but the marsh-leaping fish has long seen everything.

#1082: However, the marsh-jumping fish has long seen through all time limits:1000msSingle Point time limit:1000msMemory Limit:256MB

Describe

FJXMLHX is being brushed by a swamp fish every day, so he's eagerly finding you. You want to write a program that masks all the sentences in the Marsh Jump Fish ("marshtomp", not case sensitive). In order to make the sentence not lack of ingredients, unified replaced by "FJXMLHX".

Input

The input includes multiple lines.

Each line is a string with a length of not more than 200.

The end of a line has no relation to the beginning of the next line.

Output

The output contains multiple lines, as the input follows the result of the transformation in the description.

Sample input

The Marshtomp have seen it all before.marshtomp are beaten by fjxmlhx! Amarshtompb

Sample output

The FJXMLHX have seen it all before.fjxmlhx are beaten by fjxmlhx! Afjxmlhxb

Why do I have to do water problems again haha

#include <stdio.h> #include <string.h>int main () {char str[205];while (gets (str)!=null) {int Len=strlen (str ); for (int i=0;i<len;i++) {if (I+8>=len) break;if ((str[i]== ' m ' | | | str[i]== ' M ') && (str[i+1]== ' A ' | | str[i+1]== ' A ') && (str[i+2]== ' R ' | | str[i+2]== ' R ') && (str[i+3]== ' s ' | | str[i+3]== ' S ') && (str[i+4]== ' h ' | | str[i+4]== ' H ') && (str[i+5]== ' t ' | | str[i+5]== ' T ') && (str[i+6]== ' O ' | | str[i+6]== ' O ' && (str[i+7]== ' m ' | | str[i+7]== ' M ') && (str[i+8]== ' P ' | | str[i+8]== ' P ')) {str[i]= ' F '; str[i+1]= ' j '; str[i+2]= ' x '; str[i+3]= ' m '; str[i+4]= ' l '; str[i+5]= ' h '; str[i+6]= ' x '; str[i +7]= ' * '; str[i+8]= ' * ';}} for (int i=0;i<len;i++) {if (str[i]!= ' * ') printf ("%c", Str[i]);} printf ("\ n");}}

hiho#1082 but the marsh-leaping fish has long seen everything.