# The Love-Letter Mystery

#codeifyoucansolve
James found a love letter his friend Harry has written for his girlfriend. James is a prankster, so he decides to meddle with the letter. He changes all the words in the letter into palindromes.

To do this, he follows 2 rules:

(a) He can reduce the value of a letter, e.g. he can change ‘d’ to ‘c’, but he cannot change ‘c’ to ‘d’.
(b) In order to form a palindrome, if he has to repeatedly reduce the value of a letter, he can do it until the letter becomes ‘a’. Once a letter has been changed to ‘a’, it can no longer be changed.

Each reduction in the value of any letter is counted as a single operation. Find the minimum number of operations required to convert a given string into a palindrome.

Input Format
The first line contains an integer T, i.e., the number of test cases.
The next T lines will contain a string each. The strings do not contain any spaces.

Output Format
A single line containing the number of minimum operations corresponding to each test case.

Constraints
1 ≤ T ≤ 10
1 ≤ length of string ≤ 104
All characters are lower case English letters.

Sample Input #00

4
abc
abcba
abcd
cba

Sample Output #00

2
0
4
2

Explanation

For the first test case, abc -> abb -> aba.
For the second test case, abcba is already palindromic string.
For the third test case, abcd -> abcc -> abcb -> abca = abca -> abba.
For the fourth test case, cba -> bba -> aba.

```#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define MAX 10000
int main() {
int T,i,j,sum;
char str[MAX];
scanf("%d",&T);
while(T--)
{
scanf("%s",str);
sum =   0;
for(i=0,j=strlen(str)-1;i<j;i++,j--)
{
sum +=abs(str[i]-str[j]);
}
printf("%d\n",sum);
}
return 0;
}
```