/*
Solutie Dani Mocanu O(-1)
sol += min() - max() + 1
lucram cu abscise, in min() fixam abscisa pctului de intersectie dintre dreapta Fib[i] si cel mai la dreapta-jos pct din dreptunghi
in max() fixam abscisa pctului de intersectie dintre dreapta Fib[i] si cel mai la stanga-sus pct din dreptunghi
dreapta fib[i] = toate perechile (x,y) care x+y = fib[i]
dupa ce avem cele doua abscise fixate, facem diferenta si adaugam 1 PTR CA E INTERVAL INCHIS evident wtf
grije ca fib[i] sa se intersecteze cu dreptunghiu, altfel faci balarii
nu vad dinamica
*/
#include <fstream>
using namespace std;

const int BFSIZE = 1<<18;
char S[BFSIZE], *p, O[2000000], *o;
inline long long Get();
inline void Check();

ifstream is ("fibo3.in");
ofstream os ("fibo3.out");

long long A, B, C, D, N;
long long Fib[94];

inline void CheatOutput(long long X);

int main()
{
    p = S;
    o = O;
    Fib[0] = Fib[1] = 1;
    for (int i = 2; i < 94; ++i)
        Fib[i] = Fib[i-1] + Fib[i-2];

    N = Get();
    int i;
    for (long long Sol; N; --N)
    {
        A = Get(); B = Get(); C = Get(); D = Get();
        for (i = 1; Fib[i] < A+B; ++i);
        for (Sol = 0; Fib[i] <= C+D; ++i)
            Sol += (min(C, Fib[i]-B) - max(A, Fib[i]-D) + 1);
        CheatOutput(Sol);
    }

    os << O;
    is.close();
    os.close();
}

inline long long Get()
{
    while (*p < '0' || '9' < *p) ++p, Check();
    long long X = 0;
    while ('0' <= *p && *p <= '9') X = X * 10 + (*p-'0'), ++p, Check();
    return X;
};

inline void Check()
{
    if (*p == '\0') is.get(S, BFSIZE, '\0'), p = S;
};

inline void CheatOutput(long long X)
{
    long long Y = 0;
    for (;X; X /= 10)
        Y = Y*10 + X%10;
    for (;Y; Y/= 10, ++o)
        *o = (Y%10) + '0';
    *o = '\n'; ++o;
};