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In written languages, some symbols may appear more often than others. Expected frequency tables have been defined for many languages. For each symbol in a language, a frequency table will contain its expected percentage in a typical passage written in that language. For example, if the symbol "a" has an expected percentage of 5, then 5% of the letters in a typical passage will be "a". If a passage contains 350 letters, then 'a' has an expected count of 17.5 for that passage (17.5 = 350 * 5%). Please note that the expected count can be a non-integer value.

The deviation of a text with respect to a language frequency table can be computed in the following manner. For each letter ('a'-'z') determine the difference between the expected count and the actual count in the text. The deviation is the sum of the squares of these differences. Blank spaces (' ') and line breaks (each element of text is a line) are ignored when calculating percentages.

Each frequency table will be described as a concatenation of up to 16 strings of the form "ANN", where A is a lowercase letter ('a'-'z') and NN its expected frequency as a two-digit percentage between "00" (meaning 0%) and "99" (meaning 99%), inclusive. Any letter not appearing in a table is not expected to appear in a typical passage (0%). You are given a String[] frequencies of frequency tables of different languages. Return the lowest deviation the given text has with respect to the frequency tables.

Examples

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{"a30b30c40","a20b40c40"}

{"aa bbbb cccc"}

Returns: 0.0

The first table indicates that 30% of the letters are expected to be 'a', 30% to be 'b', and 40% to be 'c'. The second table indicates that 20% are expected to be 'a', 40% to be 'b', and 40% to be 'c'. We consider the text to have length 10, as blank spaces are ignored. With respect to the first table, there are 2 'a' where 3 were expected (a difference of 1), one more 'b' than expected (again a difference of 1) and as many 'c' as expected. The sum of...