Problem:
Score : 1000 points

Problem Statement

You have two strings A = A_1 A_2 ... A_n and B = B_1 B_2 ... B_n of the same length consisting of 0 and 1.

You can transform A using the following operations in any order and as many times as you want:

Shift A by one character to the left (i.e., if A = A_1 A_2 ... A_n, replace A with A_2 A_3 ... A_n A_1).

Shift A by one character to the right (i.e., if A = A_1 A_2 ... A_n, replace A with A_n A_1 A_2 ... A_{n-1}).

Choose any i such that B_i = 1. Flip A_i (i.e., set A_i = 1 - A_i).

You goal is to make strings A and B equal.

Print the smallest number of operations required to achieve this, or -1 if the goal is unreachable.

Constraints

1 \leq |A| = |B| \leq 2,000

A and B consist of 0 and 1.

Input

Input is given from Standard Input in the following format:

A
B

Output

Print the smallest number of operations required to make strings A and B equal, or -1 if the goal is unreachable.

Sample Input 1

1010
1100

Sample Output 1

3

Here is one fastest way to achieve the goal:

Flip A_1: A = 0010

Shift A to the left: A = 0100

Flip A_1 again: A = 1100

Sample Input 2

1
0

Sample Output 2

-1

There is no way to flip the only bit in A.

Sample Input 3

11010
10001

Sample Output 3

4

Here is one fastest way to achieve the goal:

Shift A to the right: A = 01101

Flip A_5: A = 01100

Shift A to the left: A = 11000

Shift A to the left again: A = 10001

Sample Input 4

0100100
1111111

Sample Output 4

5

Flip A_1, A_3, A_4, A_6 and A_7 in any order.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/*

This Submission is to determine how many 120/240 min const. delivery point there are.

//info
120 req. steps <= 5
*/
#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <random>
#include<map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <unordered_map>
#include <cstdlib>
#include <complex>
#include <cctype>
#include <bitset>
using namespace std;
typedef string::const_iterator State;
#define Ma_PI 3.141592653589793
#define eps 1e-5
#define LONG_INF 1e18
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007LL
#define MOD 998244353LL
#define seg_size 262144*4
#define REP(a,b) for(long long a = 0;a < b;++a)

unsigned long xor128() {
	static unsigned long x = time(NULL), y = 362436069, z = 521288629, w = 88675123;
	unsigned long t = (x ^ (x << 11));
	x = y; y = z; z = w;
	return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}
double dot(complex<double> a, complex<double> b) {
	return a.real() * b.real() + a.imag() * b.imag();
}
double gyaku_dot(complex<double> a, complex<double> b) {
	return a.real() * b.imag() - a.imag() * b.real();
}
double leng(complex<double> a) {
	return sqrt(a.real() * a.real() + a.imag() * a.imag());
}
double angles(complex<double> a, complex<double> b) {
	double cosine = dot(a, b) / (leng(a) * leng(b));
	double sine = gyaku_dot(a, b) / (leng(a) * leng(b));
	double kaku = acos(min((double)1.0, max((double)-1.0, cosine)));
	if (sine <= 0) {
		kaku = 2 * Ma_PI - kaku;
	}
	return kaku;
}
vector<int> convex_hull(vector<complex<double>> a) {
	vector<int> ans;
	double now_minnest = a[0].real();
	int now_itr = 0;
	REP(i, a.size()) {
		if (now_minnest > a[i].real()) {
			now_minnest = a[i].real();
			now_itr = i;
		}
	}
	ans.push_back(now_itr);
	complex<double> ba(0, 1);
	while (true) {
		int now_go = 0;
		double now_min = 0;
		double now_length = 0;
		int starter = ans[ans.size() - 1];
		for (int i = 0; i < a.size(); ++i) {
			if (i != starter) {
				double goa = angles(ba, a[i] - a[starter]);
				if (goa - now_min >= eps || (abs(goa - now_min) <= eps && (abs(a[i] - a[starter]) - now_length) >= eps)) {
					now_min = goa;
					now_go = i;
					now_length = abs(a[i] - a[starter]);
				}
			}
		}
		if (now_go == ans[0]) break;
		ans.push_back(now_go);
		ba = complex<double>(a[now_go] - a[starter]);
	}
	return ans;
}
int left_move_req[3000];
int right_move_req[3000];
int main(){
	string a, b;
	cin >> a >> b;
	string c = b;
	sort(c.begin(), c.end());
	if (c[c.size() - 1] == '0') {
		if (a == b) {
			cout << 0 << endl;
		}
		else {
			cout << -1 << endl;
		}
		return 0;
	}
	for (int i = 0; i < b.length(); ++i) {
		if (b[i] == '1') {
			for (int q = 1;; ++q) {
				if (b[(q + i) % b.length()] == '1') break;
				left_move_req[(q + i) % b.length()] = q;
			}
			for (int q = 1;; ++q) {
				if (b[(i-q + b.length()) % b.length()] == '1') break;
				right_move_req[(i - q + b.length()) % b.length()] = q;
			}
		}
	}
	//Welcome うさ!
	int ans = 100000;
	for (int left_move = 0; left_move < a.length(); ++left_move) {
		int right_move = 0;
		int type_c = 0;
		for (int q = 0; q < a.length(); ++q) {
			int now_itr = (q + left_move + (int)a.length()) % a.length();
			if (b[q] == '1') {
				if (a[now_itr] == '0') {
					type_c++;
				}
				continue;
			}
			else {
				if (a[now_itr] == '1') {
					//erase it
					type_c++;
					if (left_move_req[now_itr] <= left_move) {
						//ok
						continue;
					}
					else {
						right_move = max(right_move, right_move_req[now_itr]);
					}
				}
			}
		}
		ans = min(ans, right_move * 2 + left_move + type_c);
	}
	for (int right_move = 0; right_move < a.length(); ++right_move) {
		int left_move = 0;
		int type_c = 0;
		for (int q = 0; q < a.length(); ++q) {
			int now_itr = (q - right_move + (int)a.length()) % a.length();
			if (b[q] == '1') {
				if (a[now_itr] == '0') {
					type_c++;
				}
				continue;
			}
			else {
				if (a[now_itr] == '1') {
					//erase it
					type_c++;
					if (right_move_req[now_itr] <= right_move) {
						//ok
						continue;
					}
					else {
						left_move = max(left_move, left_move_req[now_itr]);
					}
				}
			}
		}
		ans = min(ans, left_move * 2 + right_move + type_c);
	}
	cout << ans << endl;
}