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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define DUMP(x) cerr << #x << "=" << x << endl
#define DUMP2(x, y) cerr<<"("<<#x<<", "<<#y<<") = ("<<x<<", "<<y<<")"<< endl
#define BINARY(x) static_cast<bitset<16> >(x)

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define REP(i,m,n) for (int i=m;i<(int)(n);i++)

#define in_range(x, y, w, h) (0<=(int)(x) && (int)(x)<(int)(w) && 0<=(int)(y) && (int)(y)<(int)(h))
#define ALL(a) (a).begin(),(a).end()

typedef long long ll;
const int INF   = 1e9;
const ll  INFLL = 1e18;
typedef pair<int, int> PII;
typedef vector<vector<int>> Graph;
int dx[4]={0, -1, 1, 0}, dy[4]={-1, 0, 0, 1};

bool check(string S)
{
    string T = S;
    reverse(ALL(T));
    return S == T;
}

int main()
{
    ios::sync_with_stdio(false);

    string S; cin >> S;
    int N = S.size();
    string T;
    for (auto c : S) if (c != 'x') T += c;

    bool ok = check(T);
    if (!ok) {
        cout << -1 << endl;
        return 0;
    }

    int l = 0, r = N-1;
    int ans = 0;
    while (l<r) {
        if (S[l] == 'x' && S[l] != S[r]) {
            ans++;
            l++;
        } else if (S[r] == 'x' && S[l] != S[r]) {
            ans++;
            r--;
        } else {
            l++, r--;
        }
    }
    cout << ans << endl;
}