CodeforcesB. Camp Schedule （KMP）

B. Camp Schedule

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

The new camp by widely-known over the country Spring Programming Camp is going to start soon. Hence, all the team of friendly curators and teachers started composing the camp's schedule. After some continuous discussion, they came up with a schedule ss, which can be represented as a binary string, in which the ii-th symbol is '1' if students will write the contest in the ii-th day and '0' if they will have a day off.

At the last moment Gleb said that the camp will be the most productive if it runs with the schedule tt (which can be described in the same format as schedule ss). Since the number of days in the current may be different from number of days in schedule tt, Gleb required that the camp's schedule must be altered so that the number of occurrences of tt in it as a substring is maximum possible. At the same time, the number of contest days and days off shouldn't change, only their order may change.

Could you rearrange the schedule in the best possible way?

Input

The first line contains string ss (1⩽|s|⩽5000001⩽|s|⩽500000), denoting the current project of the camp's schedule.

The second line contains string tt (1⩽|t|⩽5000001⩽|t|⩽500000), denoting the optimal schedule according to Gleb.

Strings ss and tt contain characters '0' and '1' only.

Output

In the only line print the schedule having the largest number of substrings equal to tt. Printed schedule should consist of characters '0' and '1' only and the number of zeros should be equal to the number of zeros in ss and the number of ones should be equal to the number of ones in ss.

In case there multiple optimal schedules, print any of them.

Examples

input

Copy

101101
110

output

Copy

110110

input

Copy

10010110
100011

output

Copy

01100011

input

Copy

10
11100

output

Copy

01

Note

In the first example there are two occurrences, one starting from first position and one starting from fourth position.

In the second example there is only one occurrence, which starts from third position. Note, that the answer is not unique. For example, if we move the first day (which is a day off) to the last position, the number of occurrences of tt wouldn't change.

In the third example it's impossible to make even a single occurrence.

题目大意：给你一个文本串和一个模式串。要求你重组文本串，构造出一个尽可能多的包含模式串的字符串。

解题思路：对文本串进行遍历，统计下1和0的个数。求出模式串的Next数组，因为根据模式串去构造的时候，需要尽可能多的重复构造模式串，所以每次构造一次模式串之后，需要回来最长公共前后缀不会点的那个位置。只需要模拟一下就可以了。

题目链接：https://codeforces.com/contest/1137/problem/B

/*
@Author: Top_Spirit
@Language: C++
*/
#include <bits/stdc++.h>
using namespace std ;
const int Maxn = 5e5 + 10 ;

string s1, s2 ;
int len1, len2 ;
int Next[Maxn] ;
string ans ;

void getNext() {
    Next[0] = -1 ;
    int k = -1, j = 0 ;
//    cout << len2 << endl ;
    while (j < len2) {
        if (k == -1 || s2[j] == s2[k]) Next[++j] = ++k ;
        else k = Next[k] ;
    }
}

int main (){
    cin >> s1 >> s2 ;
    len1 = s1.size() ;
    len2 = s2.size() ;
    getNext() ;
    int cnt0 = 0, cnt1 = 0 ;
    for (int i = 0; i < len1; i++) s1[i] == '0' ? cnt0++ : cnt1++ ;
    int j = 0 ;
    while (cnt0 > 0 && cnt1 > 0){
        if (s2[j] == '0'){
            if (cnt0 > 0) ans += "0", cnt0-- ;
            else break ;
        }
        else if (s2[j] == '1') {
            if (cnt1 > 0) ans += "1", cnt1-- ;
            else break ;
        }
        j++ ;
        if (j == len2 + 1) j = Next[len2] ;
    }
    cout << ans ;
    if (cnt0) while (cnt0--) cout << 0 ;
    if (cnt1) while (cnt1--) cout << 1 ;
    cout << endl ;
    return 0 ;
}