Problem: You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.

For example:

  * by applying a move to the string "where", the result is the string "here", 
  * by applying a move to the string "a", the result is an empty string "". 



You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.

Write a program that finds the minimum number of moves to make two given strings s and t equal.

Input

The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2⋅10^5, inclusive.

Output

Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.

Examples

Input

test
west


Output

2


Input

codeforces
yes


Output

9


Input

test
yes


Output

7


Input

b
ab


Output

1

Note

In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".

In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" → "es". The move should be applied to the string "yes" once. The result is the same string "yes" → "es".

In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.

In the fourth example, the first character of the second string should be deleted.

Solution (in JAVA):

Solution:

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

public class deletefromtheleft {
    static Calendar ts, te;                                                 //For time calculation
    static int mod9 = (int) 1e9 + 7;

    static class InputReader {                                              //InputReader class for fast input
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private SpaceCharFilter filter;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (numChars == -1)
                throw new InputMismatchException();

            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }

                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        public String nextLine() {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        public int nextInt() {
            int c = read();

            while (isSpaceChar(c))
                c = read();

            int sgn = 1;

            if (c == '-') {
                sgn = -1;
                c = read();
            }

            int res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));

            return res * sgn;
        }

        public long nextLong() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;

            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));
            return res * sgn;
        }

        public double nextDouble() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            double res = 0;
            while (!isSpaceChar(c) && c != '.') {
                if (c == 'e' || c == 'E')
                    return res * Math.pow(10, nextInt());
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            if (c == '.') {
                c = read();
                double m = 1;
                while (!isSpaceChar(c)) {
                    if (c == 'e' || c == 'E')
                        return res * Math.pow(10, nextInt());
                    if (c < '0' || c > '9')
                        throw new InputMismatchException();
                    m /= 10;
                    res += (c - '0') * m;
                    c = read();
                }
            }
            return res * sgn;
        }

        public String readString() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            }
            while (!isSpaceChar(c));

            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            if (filter != null)
                return filter.isSpaceChar(c);
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public String next() {
            return readString();
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);
        }
    }

    //static HashMap<Integer, ArrayList<Pair>> adl;

    //
    ///////Main function starts here... threading has been used to increase the stack size in case we use recursion which involves higher stack size...////////
    //

    public static void main(String[] args) {
        new Thread(null, null, "BlackRise", (long) 1e1)          //the last parameter is stack size which is desired, it is safe to chose 10^9 for all ques.
        {
            public void run() {
                Blackrise();
            }
        }.start();
    }

    static void Blackrise() {                                                       //The name Blackrise is my pen name... you can change the name according to your wish........
        InputReader input = new InputReader(System.in);
        PrintWriter pw = new PrintWriter(System.out, true);
        ts = Calendar.getInstance();
        ts.setTime(new Date());

        char ch1[]=input.readString().trim().toCharArray();
        char ch2[]=input.readString().trim().toCharArray();
        int l1=ch1.length;
        int l2=ch2.length;
        if(l1>=l2)find(ch1,ch2,l1,l2);
        else find(ch2,ch1,l2,l1);


        te = Calendar.getInstance();
        te.setTime(new Date());
        //pw.printf("\nExecution time was :- %f s\n", (te.getTimeInMillis() - ts.getTimeInMillis()) / 1000.00);
    }
    static void find(char ch1[], char ch2[], int l1,int l2)
    {
        int i;
        for(i=1;i<=l2;i++)
        {
            if(ch1[l1-i]!= ch2[l2-i])break;
        }
        if(i==1)System.out.println(l1+l2);
        else System.out.println(l1+l2 - 2*(i-1));
    }

}