CF1005B.Delete from the Left

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.

给你两个字符串 $s$ 和 $t$。在一次操作中，你可以任选其中一个字符串，并删除其第一个（即最左侧）字符。执行一次操作后，该字符串的长度减少 $1$。若某个字符串已为空，则不能再对其执行操作。

例如：

• 对字符串 "where" 执行一次操作，结果为字符串 "here"；
• 对字符串 "a" 执行一次操作，结果为空字符串 ""。

你需要通过最少的操作次数，使给定的两个字符串相等。最终两个字符串可能均为空字符串，此时它们自然相等。在这种情况下，答案显然等于两个初始字符串长度之和。

请编写一个程序，求出使两个给定字符串 $s$ 和 $t$ 相等所需的最少操作次数。

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 $2\cdot10^5$, inclusive.

输入的第一行包含字符串 $s$。输入的第二行包含字符串 $t$。两个字符串均由小写拉丁字母组成。每个字符串的字符数在 $1$ 到 $2\cdot10^5$ 之间（含端点）。

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.

输出所需的最少移动次数。最终，两个字符串都有可能均变为空字符串，从而彼此相等。此时，答案显然为给定字符串长度之和。

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" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "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.

在第一个例子中，应对第一个字符串执行一次操作，对第二个字符串也执行一次操作。结果，两个字符串都将变为 "est"。

在第二个例子中，应对字符串 "codeforces" 执行 $8$ 次操作。结果，该字符串变为 "codeforces" $\to$ "es"。应对字符串 "yes" 执行一次操作，结果为相同的字符串 "yes" $\to$ "es"。

在第三个例子中，只有将两个字符串完全删除，才能使它们相等。也就是说，最终两个字符串均为空。

在第四个例子中，应删除第二个字符串的第一个字符。