ICPC NEERC 2010 B - Binary Operation

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Points: 20 (partial)
Time limit: 1.0s
Memory limit: 64M

Problem type

Consider a binary operation \odot defined on digits 0 to 9, \odot : \{0, 1, \dots, 9\} \times \{0, 1, \dots, 9\} \to \{0, 1, \dots, 9\}, such that 0 \odot 0 = 0.

A binary operator \otimes is a generalization of \odot to the set of non-negative integers, \otimes : \mathbb Z_{0+} \times \mathbb Z_{0+} \to \mathbb Z_{0+}. The result of a \otimes b is defined in the following way: if one of the numbers a and b has fewer digits than the other in decimal notation, then append leading zeroes to it, so that the numbers are of the same length; then apply the operation \odot digit-wise to the corresponding digits of a and b.

Example. If a \odot b = ab \bmod 10, then 5566 \otimes 239 = 84.

Let us define \otimes to be left-associative, that is, a \otimes b \otimes c is to be interpreted as (a \otimes b) \otimes c.

Given a binary operation \odot and two non-negative integers a and b, calculate the value of a \otimes (a+1) \otimes (a+2) \otimes \dots \otimes (b-1) \otimes b.

Input Specification

The first ten lines of the input contain the description of the binary operation \odot. The i^\text{th} line of the input contains a space-separated list of ten digits - the j^\text{th} digit in this list is equal to (i-1) \odot (j-1).
The first digit in the first line is always 0.
The eleventh line of the input contains two non-negative integers a and b (0 \le a \le b \le 10^{18}).

Output Specification

Output a single number - the value of a \otimes (a+1) \otimes (a+2) \otimes \dots \otimes (b-1) \otimes b without extra leading zeroes.

Sample Input

0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 0
2 3 4 5 6 7 8 9 0 1
3 4 5 6 7 8 9 0 1 2
4 5 6 7 8 9 0 1 2 3
5 6 7 8 9 0 1 2 3 4
6 7 8 9 0 1 2 3 4 5
7 8 9 0 1 2 3 4 5 6
8 9 0 1 2 3 4 5 6 7
9 0 1 2 3 4 5 6 7 8
0 10

Sample Output

15

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