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 Python (3.4.3) (judge verdict: Runtime Error)

# C


S = input()
N = len(S)
# 回分判定
ss = ""
for s in S:
    if s == 'x':
        pass
    else:
        ss += s
if ss != ss[::-1]:
    print(-1)
    exit()

# 真ん中を決める
# x の位置と個数を求める
n = len(ss)
if n % 2 == 0:
    cnt = 0
    for i in range(n//2+1):
        for j in range(cnt, N):
            if S[j] == "x":
                continue
            else:
                cnt = j+1
                break
        if i == n//2-1:
            left_center = cnt-1
    right_center = cnt-1
    # 真ん中の分を先に計算する
    ans = (abs(right_center - left_center) % 2)
    # centerから両側に展開していく
    count = 0
    x_left = []
    x_right = []
    for i in range(left_center):
        if S[left_center-1 - i] == 'x':
            count += 1
        else:
            x_left.append(count)
            count = 0
    x_left.append(count)
    count = 0
    for i in range(right_center+1, N, 1):
        if S[i] == 'x':
            count += 1
        else:
            x_right.append(count)
            count = 0
    x_right.append(count)
    ans = 0
    for i in range(len(x_left)):
        ans += abs(x_left[i] - x_right[i])
    print(ans)
else:
    cnt = 0
    for i in range(n//2+1):
        for j in range(cnt, N):
            if S[j] == "x":
                continue
            else:       # S[j] != "x" なら ss[i] と S[j] は一致する
                cnt = j + 1
                break
    center = cnt-1
    # print(center)
    # center から両側に展開していき，center からの位置と x の個数を求める
    distans = 0     # x 以外の文字が出るたびに更新する
    count = 0
    x_left = []
    x_right = []
    for i in range(center):
        if S[center-1 - i] == "x":
            count += 1
        else:
            x_left.append([distans, count])
            distans += 1
            count = 0
    x_left.append([distans, count])
    distans = 0
    count = 0
    for i in range(center+1, N):
        if S[i] == "x":
            count += 1
        else:
            x_right.append([distans, count])
            distans += 1
            count = 0
    x_right.append([distans, count])
    # print(x_left)
    # print(x_right)
    ans = 0
    for i in range(len(x_left)):
        ans += abs(x_left[i][1] - x_right[i][1])
    print(ans)