Problem:
Score : 400 points

Problem Statement

We have a string s consisting of lowercase English letters.
Snuke can perform the following operation repeatedly:

Insert a letter x to any position in s of his choice, including the beginning and end of s.

Snuke's objective is to turn s into a palindrome.
Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.

Notes

A palindrome is a string that reads the same forward and backward.
For example, a, aa, abba and abcba are palindromes, while ab, abab and abcda are not.

Constraints

1 \leq |s| \leq 10^5

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

If the objective is achievable, print the number of operations required.
If it is not, print -1 instead.

Sample Input 1

xabxa

Sample Output 1

2

One solution is as follows (newly inserted x are shown in bold):

xabxa → xaxbxa → xaxbxax

Sample Input 2

ab

Sample Output 2

-1

No sequence of operations can turn s into a palindrome.

Sample Input 3

a

Sample Output 3

0

s is a palindrome already at the beginning.

Sample Input 4

oxxx

Sample Output 4

3

One solution is as follows:

oxxx → xoxxx → xxoxxx → xxxoxxx

Solution in Kotlin (1.3.71) (judge verdict: Accepted)

import kotlin.math.abs

fun main() {
    val s = readLine() ?: return
    fun String.isPalindrome() = take(length / 2) == takeLast(length / 2).reversed()
    val withoutX = s.replace("x", "")
    if (!withoutX.isPalindrome()) {
        println("-1")
        return
    }
    var i = 0
    var index = 0
    val half = withoutX.length / 2 + withoutX.length % 2
    val counter = Array(half) { 0 }
    while (half > i) {
        when (s[index]) {
            'x' -> counter[i]++
            else -> {
                i++
            }
        }
        index++
    }
    i = 0
    index = 0
    var count = 0
    while (half > i) {
        when (s[s.length - index - 1]) {
            'x' -> counter[i]--
            else -> {
                count += abs(counter[i])
                i++
            }
        }
        index++
    }
    println(count.toString())
}