The program must accept a string S containing only alphabets as the input. The program must print the string S for (L*2)-1 times (where L is the length of the string S) based on the following conditions.
– In line 1, the program must print L-1 hyphens and the first alphabet in S.
– In line 2, the program must print L-2 hyphens and the first two alphabets in S.
– Similarly, the program prints the first L lines as the output.
– In line L+1, the program must print the string S except the first alphabet.
– In line L+2, the program must print the string S except the first two alphabets.
– Similarly, the program prints the remaining lines as the output.

Boundary Condition(s):
3 <= Length of S <= 100

Input Format:
The first line contains S.

Output Format:
The first (L*2)-1 lines containing the string values based on the given conditions.

Example Input/Output 1:
Input:
receiving

Output:
——–r
——-re
——rec
—–rece
—-recei
—receiv
–receivi
-receivin
receiving
eceiving
ceiving
eiving
iving
ving
ing
ng
g

Explanation:
The length of the given string receiving is 9. So the pattern contains (9*2)-1 lines.
In line 1, 8 hyphens and the first alphabet in S are printed.
In line 2, 7 hyphens and the first two alphabets in S are printed.
In line 3, 6 hyphens and the first three alphabets in S are printed.
Similarly, the first 9 lines are printed.
——–r
——-re
——rec
—–rece
—-recei
—receiv
–receivi
-receivin
receiving
In line 10, the string S is printed except the first alphabet.
In line 11, the string S is printed except the first two alphabets.
In line 12, the string S is printed except the first three alphabets.
Similarly, the remaining lines are printed.
eceiving
ceiving
eiving
iving
ving
ing
ng
g

Example Input/Output 2:
Input:
ZERO

Output:
—Z
–ZE
-ZER
ZERO
ERO
RO
O

Example Input/Output 3:
Input:
KeyBoard

Output:
——-K
——Ke
—–Key
—-KeyB
—KeyBo
–KeyBoa
-KeyBoar
KeyBoard
eyBoard
yBoard
Board
oard
ard
rd
d

C:

#include<stdio.h>
#include <stdlib.h>
int main()
{
int ele,foo,num;
char strin[101];
scanf("%strin",strin);
for(ele=0;ele<strlen(strin);ele++)
{
    for(foo=0;foo<strlen(strin)-ele-1;foo++)
    printf("-");
    for(foo=0;foo<=ele;foo++)
    printf("%c",strin[foo]);
    printf("n");
}
for(ele=1;ele<strlen(strin);ele++)
{
    printf("%strin",&strin[ele]);
    printf("n");
}
}

C:

#include<stdio.h>
#include <stdlib.h>
int main()
{
    char str[10001];
    scanf("%strin",str);
    int length=strlen(str);
    for(int ele=0;ele<(length*2)-1;ele++){
        for(int foo=0;foo<length-ele-1;foo++){
            printf("-");
        }
        int var1,var2;
if(length-ele-1>=0){
var1=0;
}
else{
var1=ele%length+1;
}
if(length-ele-1>=0){
var2=ele+1;
}
else{
var2=length;
}
        for(int foo=var1;foo<var2;foo++){
printf("%c",str[foo]);
}
        printf("n");
    }
    return 1;
}