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.util.*;
import java.io.*; 
import java.math.BigInteger;
import java.text.DecimalFormat;

public class Main{
    final long mod = (int)1e9+7, IINF = (long)1e19;
    final int MAX = (int)1e6+2, MX = (int)1e7+1, INF = (int)1e9, root = 3;
    DecimalFormat df = new DecimalFormat("0.0000000000000");
    double eps = 1e-9, PI = 3.141592653589793238462643383279502884197169399375105820974944;
    static boolean multipleTC = false, memory = false;
    FastReader in;PrintWriter out;
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }

    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        for(int i = 1, T= (multipleTC)?ni():1; i<= T; i++)solve(i);
        out.flush();
        out.close();
    }
    
    void solve(int TC) throws Exception{
        StringBuilder a = new StringBuilder(n()).reverse(), b = new StringBuilder(n()).reverse();
        int ans = a.length()+b.length();
        for(int i = 0; i< Math.min(a.length(), b.length()); i++)if(a.charAt(i)==b.charAt(i))ans-=2;else break;
        pn(ans);
    }

    int[] sort(int[] a){
        if(a.length==1)return a;
        int mid = a.length/2;
        int[] b = sort(Arrays.copyOfRange(a,0,mid)), c = sort(Arrays.copyOfRange(a,mid,a.length));
        for(int i = 0, j = 0, k = 0; i< a.length; i++){
            if(j<b.length && k<c.length){
                if(b[j]<c[k])a[i] = b[j++];
                else a[i] = c[k++];
            }else if(j<b.length)a[i] = b[j++];
            else a[i] = c[k++];
        }
        return a;
    }

    long[] sort(long[] a){
        if(a.length==1)return a;
        int mid = a.length/2;
        long[] b = sort(Arrays.copyOfRange(a,0,mid)), c = sort(Arrays.copyOfRange(a,mid,a.length));
        for(int i = 0, j = 0, k = 0; i< a.length; i++){
            if(j<b.length && k<c.length){
                if(b[j]<c[k])a[i] = b[j++];
                else a[i] = c[k++];
            }else if(j<b.length)a[i] = b[j++];
            else a[i] = c[k++];
        }
        return a;
    }

    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{    
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }   
            return str;
        }
    }
}