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 final class TaskB
{
	public static void main(String[] args)
	{
		new TaskB(System.in, System.out);
	}

	static class Solver implements Runnable
	{
        String s, t;
		BufferedReader in;
        PrintWriter out;

		void solve() throws IOException
		{
			s = in.readLine();
			t = in.readLine();

			int cnt = 0;
			int l = 0;
			int r = 0;

			while (s.length() - l > t.length())
				l++;

			while (t.length() - r > s.length())
				r++;

			cnt += l + r;
			s = s.substring(l, s.length());
			t = t.substring(r, t.length());

//			System.out.println("s : " + s + ", t : " + t);

			int len = s.length();
			int pos = -1;

			for (int i = len - 1; i >= 0; i--)
			{
				if (s.charAt(i) != t.charAt(i))
				{
					pos = i;

					break;
				}
			}

			for (int i = 0; i <= pos; i++)
				cnt += 2;

			out.println(cnt);
		}

		void debug(Object... o)
		{
			System.err.println(Arrays.deepToString(o));
		}

		public Solver(BufferedReader in, PrintWriter out)
		{
			this.in = in;
			this.out = out;
		}

		@Override
		public void run()
		{
			try
			{
				solve();
			}
			catch (IOException e)
			{
				e.printStackTrace();
			}
		}

	}

	static class CMath
	{
		static long power(long number, long power)
		{
			if (number == 1 || number == 0 || power == 0)
				return 1;

			if (power == 1)
				return number;

			if (power % 2 == 0)
				return power(number * number, power / 2);
			else
				return power(number * number, power / 2) * number;
		}

		static long modPower(long number, long power, long mod)
		{
			if (number == 1 || number == 0 || power == 0)
				return 1;

			number = mod(number, mod);

			if (power == 1)
				return number;

			long square = mod(number * number, mod);

			if (power % 2 == 0)
				return modPower(square, power / 2, mod);
			else
				return mod(modPower(square, power / 2, mod) * number, mod);
		}

		static long moduloInverse(long number, long mod)
		{
			return modPower(number, mod - 2, mod);
		}

		static long mod(long number, long mod)
		{
			return number - (number / mod) * mod;
		}

		static int gcd(int a, int b)
		{
			if (b == 0)
				return a;
			else
				return gcd(b, a % b);
		}

		static long min(long... arr)
		{
			long min = arr[0];

			for (int i = 1; i < arr.length; i++)
				min = Math.min(min, arr[i]);

			return min;
		}

		static long max(long... arr)
		{
			long max = arr[0];

			for (int i = 1; i < arr.length; i++)
				max = Math.max(max, arr[i]);

			return max;
		}

		static int min(int... arr)
		{
			int min = arr[0];

			for (int i = 1; i < arr.length; i++)
				min = Math.min(min, arr[i]);

			return min;
		}

		static int max(int... arr)
		{
			int max = arr[0];

			for (int i = 1; i < arr.length; i++)
				max = Math.max(max, arr[i]);

			return max;
		}

	}

	static class Utils
	{
		static boolean nextPermutation(int[] arr)
		{
			for (int a = arr.length - 2; a >= 0; --a)
			{
				if (arr[a] < arr[a + 1])
				{
					for (int b = arr.length - 1; ; --b)
					{
						if (arr[b] > arr[a])
						{
							int t = arr[a];

							arr[a] = arr[b];
							arr[b] = t;

							for (++a, b = arr.length - 1; a < b; ++a, --b)
							{
								t = arr[a];
								arr[a] = arr[b];
								arr[b] = t;
							}

							return true;
						}
					}
				}
			}

			return false;
		}

	}

	public TaskB(InputStream inputStream, OutputStream outputStream)
	{
		BufferedReader in = new BufferedReader(new InputStreamReader(inputStream));
		PrintWriter out = new PrintWriter(outputStream);
		Thread thread = new Thread(null, new Solver(in, out), "TaskB", 1 << 29);

		try
		{
			thread.start();
			thread.join();
		}
		catch (InterruptedException e)
		{
			e.printStackTrace();
		}
		finally
		{
			try
			{
				in.close();
			}
			catch (IOException e)
			{
				e.printStackTrace();
			}

			out.flush();
			out.close();
		}
	}

}