#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.

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#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; }