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Count the number of different ways of climbing to the top of a ladder.

You have to climb up a ladder. The ladder has exactly N rungs, numbered from 1 to N. With each step, you can ascend by one or two rungs. More precisely:

• with your first step you can stand on rung 1 or 2,
• if you are on rung K, you can move to rungs K + 1 or K + 2,
• finally you have to stand on rung N.

Your task is to count the number of different ways of climbing to the top of the ladder.

For example, given N = 4, you have five different ways of climbing, ascending by:

• 1, 1, 1 and 1 rung,
• 1, 1 and 2 rungs,
• 1, 2 and 1 rung,
• 2, 1 and 1 rungs, and
• 2 and 2 rungs.

Given N = 5, you have eight different ways of climbing, ascending by:

• 1, 1, 1, 1 and 1 rung,
• 1, 1, 1 and 2 rungs,
• 1, 1, 2 and 1 rung,
• 1, 2, 1 and 1 rung,
• 1, 2 and 2 rungs,
• 2, 1, 1 and 1 rungs,
• 2, 1 and 2 rungs, and
• 2, 2 and 1 rung.

The number of different ways can be very large, so it is sufficient to return the result modulo 2P, for a given integer P.

Write a function:

class Solution { public int[] solution(int[] A, int[] B); }

that, given two non-empty arrays A and B of L integers, returns an array consisting of L integers specifying the consecutive answers; position I should contain the number of different ways of climbing the ladder with A[I] rungs modulo 2B[I].

For example, given L = 5 and:

A = 4 B = 3 A = 4 B = 2 A = 5 B = 4 A = 5 B = 3 A = 1 B = 1

the function should return the sequence [5, 1, 8, 0, 1], as explained above.

Write an efficient algorithm for the following assumptions:

• L is an integer within the range [1..50,000];
• each element of array A is an integer within the range [1..L];
• each element of array B is an integer within the range [1..30].