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:
import java.io.*;
import java.util.*;

public class CodeForces implements Cloneable{
    /*****************************************WRITING CODE STARTS HERE******************************************/

    public static void solve(int tCase)throws IOException {
        char[] s = sc.next().toCharArray();
        char[] t = sc.next().toCharArray();
        int n = s.length;
        int m = t.length;
        int countOne = 0;
        for(char c : s)if(c=='1')countOne++;
        int countZero = n - countOne;
        StringBuilder ans = new StringBuilder();
        for(char c : t){
            if(c == '0')countZero--;
            else countOne--;
            if(countOne<0 || countZero<0){  //  not possible to form a t in s.
                out.println(String.valueOf(s));
                return;
            }
            ans.append(c);
        }
        int start = findStartIndex(t,m);
        int ind = start;
        while (true){
            if(t[ind]=='0')countZero--;
            else countOne--;
            if(countOne<0 || countZero<0)break;
            ans.append(t[ind]);
            ind++;
            if(ind == m)ind = start;
        }
        while (countOne-->0)ans.append('1');
        while (countZero-->0)ans.append('0');
        out.println(ans);
    }
    private static int findStartIndex(char[] str,int m){
       int l = 0;
       int r = 1;
       int[] lps = new int[m];
       while (r<m){
           if(str[l] != str[r]){
               if(l!=0)l = lps[l-1];
               else r++;
           }else {
               lps[r] = 1 + l;
               l++;
               r++;
           }
       }
//       out.println(lps[m-1]);
       return lps[m-1];

    }

    /*****************************************WRITING CODE ENDS HERE******************************************/

    public static void main(String[] args) throws IOException{
        int testCase = 1;
//        testCase = sc.nextInt();
        for (int i = 1; i <= testCase; i++) solve(i);
        closeIO();
    }
    static FastestReader sc = new FastestReader();              // FASTEST INPUT
    static PrintWriter out = new PrintWriter(System.out);                // FAST OUTPUT


    /**************************************HELPER FUNCTION STARTS HERE ************************************/

    public static int mod = (int) 1e9 +7;
    public static  int inf_int = (int) 1e9;
    public static  long inf_long = (long) 1e18;

    public static final String YES = "YES";
    public static final String NO = "NO";

    // euclidean algorithm time O(max (loga ,logb))
    public static long _gcd(long a, long b) {
        long x = Math.min(a, b);
        long y = Math.max(a, b);
        if (y % x == 0) return x;
        return _gcd(y % x, x);
    }

    public static long _lcm(long a, long b) {
        // lcm(a,b) * gcd(a,b) = a * b
        return (a / _gcd(a, b)) * b;
    }
    // binary exponentiation time O(logn)
    public static long _power(long x, long n) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans *= x;
                ans %= mod;
                n--;
            } else {
                x *= x;
                x %= mod;
                n >>= 1;
            }
        }
        return ans;
    }

    public static long _firstDivisor(long n) {
        if (n == 1 || n == 0) return n;
        for (long i = 2; i * i <= n; i++)
            if (n % i == 0) return i;
        return -1;
    }


    /**************************************HELPER FUNCTION ENDS HERE ************************************/

    /*****************************************FAST INPUT STARTS HERE *************************************/


    public static void closeIO() throws IOException{
        out.flush();
        out.close();
        sc.close();
    }
    private static class FastestReader {
        private static final int BUFFER_SIZE = 1 << 16;
        private final DataInputStream din;
        private final byte[] buffer;
        private int bufferPointer, bytesRead;

        public FastestReader() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public FastestReader(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }

        private int skip() throws IOException {
            int b;
            //noinspection StatementWithEmptyBody
            while ((b = read()) != -1 && isSpaceChar(b)) {}
            return b;
        }

        public String next() throws IOException {
            int b = skip();
            final StringBuilder sb = new StringBuilder();
            while (!isSpaceChar(b)) { // when nextLine, (isSpaceChar(b) && b != ' ')
                sb.appendCodePoint(b);
                b = read();
            }
            return sb.toString();
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ') {
                c = read();
            }
            final boolean neg = c == '-';
            if (neg) { c = read(); }
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg) { return -ret; }
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ') { c = read(); }
            final boolean neg = c == '-';
            if (neg) { c = read(); }
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg) { return -ret; }
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ') { c = read(); }
            final boolean neg = c == '-';
            if (neg) { c = read(); }

            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg) { return -ret; }
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) { buffer[0] = -1; }
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead) { fillBuffer(); }
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            din.close();
        }

    }
    /*****************************************FAST INPUT ENDS HERE *************************************/
}