问题3048--Fibonacci Check-up

3048: Fibonacci Check-up

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时间限制 : 1.000 sec  内存限制 : 64 MB

题目描述

Every ALPC has his own alpc-number just like alpc12, alpc55, alpc62 etc. As more and more fresh man join us. How to number them? And how to avoid their alpc-number conflicted? Of course, we can number them one by one, but that’s too bored! So ALPCs use another method called Fibonacci Check-up in spite of collision. First you should multiply all digit of your studying number to get a number n (maybe huge). Then use Fibonacci Check-up! Fibonacci sequence is well-known to everyone. People define Fibonacci sequence as follows: F(0) = 0, F(1) = 1. F(n) = F(n-1) + F(n-2), n >= 2. It’s easy for us to calculate F(n) mod m. But in this method we make the problem has more challenge. We calculate the formula , is the combination number. The answer mod m (the total number of alpc team members) is just your alpc-number

输入

First line is the testcase T. Following T lines, each line is two integers n, m ( 0 <= n <= 10^9, 1 <= m <= 30000 )

输出

Output the alpc-number.

样例输入 Copy

2
1 30000
2 30000

样例输出 Copy

1
3