Xtreme gcd sum

You are given the integer constants $a_1, b_1, \dots, a_n, b_n$. Run the code below to the end and find the value that is finally stored in sum.

sum = 0;
for (x1 = a1; x1 <= b1; x1++)
    for (x2 = a2; x2 <= b2; x2++)
        ...
            for (xn = an; xn <= bn; xn++)
                sum = sum + gcd(x1, x2, ..., xn);

gcd returns the greatest common divisor of its arguments $x_1, x_2, \dots, x_n$, and sum is a variable that holds an arbitrarily large integer.

Looks too easy? I think so too.

The first line contains a natural number $n$.

The $i$-th of the next $n$ lines contains $a_i$ and $b_i$, separated by a space.

$1 \le n \le 10\,000$ and $1 \le a_i \le b_i \le 1\,000\,000$.

C and C++ have no data type that holds an arbitrarily large integer, so print the value of sum modulo $1\,000\,000\,007$ on one line.