/**
     题意：给一个A串，一个B串，然后str[i] = str[i-1]+str[i-2];
            求解对于第n个串的第前m位最长的LBorder，LBorder 是指s[1:i]=s[n−i+1:n],
    做法：画几个串可以找到规律，比如第九个串
        a    b a   a b a    b a a b a   a b a b a a b  a    b a a  b  a a  b  a   b a  a  b  a
        [0] [0 1] [1 2 3]  [2 3 4 5 6] [4 5 6 7 8 9 10 11] [7 8 9 10 11 12 13 14 15 16 17 18 19]
        Java 
 **/
import java.util.*;
import java.math.*;

public class Main 
{
    public static void main(String args[]) 
    {
        BigInteger AA[] = new BigInteger [1000+5];
        BigInteger A[] = new BigInteger [1000+5];
        A[1] = BigInteger.ONE;
        A[2] = BigInteger.valueOf(2);
        BigInteger MOD = BigInteger.valueOf(258280327);
        BigInteger mm = BigInteger.valueOf(1);
        for (int i = 3; i <= 1001; i++) {
            A[i] = A[i-1].add(A[i-2]);
        }
        for(int i=2;i<=1001;i++)
        {
            A[i] = A[i].add(A[i-1]);
        }
        AA[1] = BigInteger.ZERO;
        AA[2] = BigInteger.ZERO;
        for(int i=3;i<=1001;i++)
        {
            AA[i] = AA[i-1].add(AA[i-2]).add(mm);
        }
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while (t-- > 0) 
        {
            int n = in.nextInt();
            BigInteger m = in.nextBigInteger();
            if (m.equals(BigInteger.valueOf(1)) || m.equals(BigInteger.valueOf(2)))
            {
                System.out.println(0);
                continue;
            }
            boolean ok = false;
            int tt = 0;
            for (int i = 1; i <= 1001; i++) 
            {
              int yy = A[i].compareTo(m);
              if(yy == 1) break;
              tt = i;
            }
            if(A[tt].compareTo(m) == 0)
            {
                 BigInteger temp = BigInteger.ZERO;
                 BigInteger res = A[tt].subtract(A[tt-1]);
                 BigInteger tmp  = AA[tt].add(res) ;
                 tmp = tmp.subtract(mm);
                 System.out.println(tmp.mod(MOD));
            }
            else 
            {
                BigInteger temp = BigInteger.ZERO;
                BigInteger res = m.subtract(A[tt]);
                BigInteger tmp  = AA[tt+1].add(res);
                tmp = tmp.subtract(mm);
                System.out.println(tmp.mod(MOD));
            }
          }   
    }
}