CF2121E Sponsor of Your Problems

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) $ .

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 < 10^9 $ ). It is guaranteed that the numbers $ l $ and $ r $ have the same length in decimal representation and do not have leading zeros.

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

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 $ .