HDU 5351 MZL's Border

Problem Description

As is known to all, MZL is an extraordinarily lovely girl. One day, MZL was playing with her favorite data structure, strings.

MZL is really like Fibonacci Sequence,
so she defines Fibonacci Strings in
the similar way. The definition of Fibonacci Strings is
given below.



1) fib1=b



2) fib2=a



3) fibi=fibi−1fibi−2, i>2



For instance, fib3=ab, fib4=aba, fib5=abaab.

Assume that a string s whose
length is n is s1s2s3...sn.
Then sisi+1si+2si+3...sj is
called as a substring of s,
which is written as s[i:j].

Assume that i<n.
If s[1:i]=s[n−i+1:n],
then s[1:i] is
called as a Border of s.
In Borders of s,
the longest Border is
called as s' LBorder.
Moreover, s[1:i]'s LBorder is
called as LBorderi.

Now you are given 2 numbers n and m.
MZL wonders what LBorderm of fibn is.
For the number can be very big, you should just output the number modulo 258280327(=2×317+1).

Note that 1≤T≤100, 1≤n≤103, 1≤m≤|fibn|.



Input

The first line of the input is a number T,
which means the number of test cases.

Then for the following T lines,
each has two positive integers n and m,
whose meanings are described in the description.



Output

The output consists of T lines.
Each has one number, meaning fibn's LBorderm modulo 258280327(=2×317+1).



Sample Input

2
4 3
5 5

Sample Output

1
2
找找规律，然后模拟一下,注意有大数
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Main
{
    public static void main(String args[])
    {
        Scanner cin = new Scanner(System.in);    
        BigInteger f[][] = new BigInteger[1005][3];
        int i;
        f[0][1]=BigInteger.ZERO;
        f[0][2]=BigInteger.ZERO;
        f[1][1]=BigInteger.ZERO;
        f[1][2]=BigInteger.ZERO;
        f[0][0]=BigInteger.ONE;
        f[1][0]=BigInteger.ONE;
        for (i=2;i<=1000;i++)
        {
            f[i][0]=f[i-1][0].add(f[i-2][0]);
            f[i][1]=f[i-1][2].subtract(f[i-1][1]);
            f[i][2]=f[i][1].add(f[i][0]).subtract(BigInteger.ONE);
        }
        int T=cin.nextInt();
        for(;T>=1;T--)
        {
            int n;
            BigInteger m;
            n=cin.nextInt();
            m=cin.nextBigInteger();
            for(i=1;i<=1000;i++)
            {
              if(m.compareTo(f[i][0])<1)
              {
                  m=f[i][1].add(m).subtract(BigInteger.ONE);
                  m=m.mod(BigInteger.valueOf(258280327));
                  System.out.println(m);
                  break;
              }
              else
                  m=m.subtract(f[i][0]);
            }
        }
    }
}