Problem: 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 s, which can be represented as a binary string, in which the i-th symbol is '1' if students will write the contest in the i-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 t (which can be described in the same format as schedule s). Since the number of days in the current may be different from number of days in schedule t, Gleb required that the camp's schedule must be altered so that the number of occurrences of t 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 s (1 ⩽ |s| ⩽ 500 000), denoting the current project of the camp's schedule.

The second line contains string t (1 ⩽ |t| ⩽ 500 000), denoting the optimal schedule according to Gleb.

Strings s and t contain characters '0' and '1' only.

Output

In the only line print the schedule having the largest number of substrings equal to t. Printed schedule should consist of characters '0' and '1' only and the number of zeros should be equal to the number of zeros in s and the number of ones should be equal to the number of ones in s.

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

Examples

Input


101101
110


Output


110110

Input


10010110
100011


Output


01100011


Input


10
11100


Output


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 t wouldn't change.

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

Solution (in JAVA):

Solution:
/**
 * @author derrick20
 */
import java.io.*;
import java.util.*;

public class CampSchedule {
    public static void main(String args[]) throws Exception {
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        String s = sc.next();
        String t = sc.next();
        // we need the earliest position it matches with it's own suffix
        int[] prefixTable = new int[t.length()];
        int j = 0; // current matched to position
        for (int i = 1; i < t.length(); i++) {
            while (j > 0 && t.charAt(j) != t.charAt(i)) {
                // we need to hop downwards while we are different
                j = prefixTable[j - 1];
                // this will repeatedly cause us to land at the char RIGHT
                // after each set of working prefixes. So, it'll let us
                // keep trying to build onto smaller prefixes (Eventually hitting 0 length
                // meaning we start from scratch again)

            }
            if (t.charAt(j) == t.charAt(i)) {
                j++;
                prefixTable[i] = j;
            }
        }

        int prefixLen = prefixTable[t.length() - 1];
        // so, we only need t from i = prefixLen to the end
        String pattern = t.substring(prefixLen);

        int[] sCt = count(s);
        int[] tCt = count(t);
        int[] pCt = count(pattern);


        StringBuilder sb = new StringBuilder();
        if (sCt[0] >= tCt[0] && sCt[1] >= tCt[1]) {
            sCt[0] -= tCt[0];
            sCt[1] -= tCt[1];
            sb.append(t);
        }
        while (sCt[0] >= pCt[0] && sCt[1] >= pCt[1]) {
            sCt[0] -= pCt[0];
            sCt[1] -= pCt[1];
            sb.append(pattern);
        }
        for (int i = 0; i < sCt[0]; i++) {
            sb.append(0);
        }
        for (int i = 0; i < sCt[1]; i++) {
            sb.append(1);
        }
        out.println(sb);
        out.close();
    }

    static int[] count(String s) {
        int[] res = new int[2];
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '0') {
                res[0]++;
            }
            else {
                res[1]++;
            }
        }
        return res;
    }

    static class FastScanner {
        public int BS = 1<<16;
        public char NC = (char)0;
        byte[] buf = new byte[BS];
        int bId = 0, size = 0;
        char c = NC;
        double cnt = 1;
        BufferedInputStream in;

        public FastScanner() {
            in = new BufferedInputStream(System.in, BS);
        }

        public FastScanner(String s) {
            try {
                in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
            }
            catch (Exception e) {
                in = new BufferedInputStream(System.in, BS);
            }
        }

        private char getChar(){
            while(bId==size) {
                try {
                    size = in.read(buf);
                }catch(Exception e) {
                    return NC;
                }
                if(size==-1)return NC;
                bId=0;
            }
            return (char)buf[bId++];
        }

        public int nextInt() {
            return (int)nextLong();
        }

        public int[] nextInts(int N) {
            int[] res = new int[N];
            for (int i = 0; i < N; i++) {
                res[i] = (int) nextLong();
            }
            return res;
        }

        public long[] nextLongs(int N) {
            long[] res = new long[N];
            for (int i = 0; i < N; i++) {
                res[i] = nextLong();
            }
            return res;
        }

        public long nextLong() {
            cnt=1;
            boolean neg = false;
            if(c==NC)c=getChar();
            for(;(c<'0' || c>'9'); c = getChar()) {
                if(c=='-')neg=true;
            }
            long res = 0;
            for(; c>='0' && c <='9'; c=getChar()) {
                res = (res<<3)+(res<<1)+c-'0';
                cnt*=10;
            }
            return neg?-res:res;
        }

        public double nextDouble() {
            double cur = nextLong();
            return c!='.' ? cur:cur+nextLong()/cnt;
        }

        public String next() {
            StringBuilder res = new StringBuilder();
            while(c<=32)c=getChar();
            while(c>32) {
                res.append(c);
                c=getChar();
            }
            return res.toString();
        }

        public String nextLine() {
            StringBuilder res = new StringBuilder();
            while(c<=32)c=getChar();
            while(c!='\n') {
                res.append(c);
                c=getChar();
            }
            return res.toString();
        }

        public boolean hasNext() {
            if(c>32)return true;
            while(true) {
                c=getChar();
                if(c==NC)return false;
                else if(c>32)return true;
            }
        }
    }
}