For two integers $$$a$$$ and $$$b$$$, we define $$$f(a, b)$$$ as the number of positions in the decimal representation of the numbers $$$a$$$ and $$$b$$$ where their digits are the same. For example, $$$f(12, 21) = 0$$$, $$$f(31, 37) = 1$$$, $$$f(19891, 18981) = 2$$$, $$$f(54321, 24361) = 3$$$.

You are given two integers $$$l$$$ and $$$r$$$ of the same length in decimal representation. Consider all integers $$$l \leq x \leq r$$$. Your task is to find the minimum value of $$$f(l, x) + f(x, r)$$$.

Input

Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows.

Each test case consists of a single line containing two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \lt 10^9$$$).

It is guaranteed that the numbers $$$l$$$ and $$$r$$$ have the same length in decimal representation and do not have leading zeros.

Output

For each test case, output the minimum value of $$$f(l, x) + f(x, r)$$$ among all integer values $$$l \leq x \leq r$$$.

Note

In the first test case, you can choose $$$x = 1$$$. Then $$$f(1, 1) + f(1, 1) = 1 + 1 = 2$$$.

In the second test case, you can choose $$$x = 2$$$. Then $$$f(2, 2) + f(2, 3) = 1 + 0 = 1$$$.

In the third test case, you can choose $$$x = 5$$$. Then $$$f(4, 5) + f(5, 6) = 0 + 0 = 0$$$.

In the fourth test case, you can choose $$$x = 15$$$. Then $$$f(15, 15) + f(15, 16) = 2 + 1 = 3$$$.

In the fifth test case, you can choose $$$x = 18$$$. Then $$$f(17, 18) + f(18, 19) = 1 + 1 = 2$$$.

In the sixth test case, you can choose $$$x = 200$$$. Then $$$f(199, 200) + f(200, 201) = 0 + 2 = 2$$$.

In the seventh test case, you can choose $$$x = 900$$$. Then $$$f(899, 900) + f(900, 999) = 0 + 1 = 1$$$.

In the eighth test case, you can choose $$$x = 1992$$$. Then $$$f(1990, 1992) + f(1992, 2001) = 3 + 0 = 3$$$.

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