Euler Problem 27: Quadratic primes

Euler discovered the remarkable quadratic formula:

n² + n + 41

It turns out that the formula will produce 40 primes for the consecutive
values n = 0 to 39. However, when n = 40, 40² + 40 + 41 = 40(40 + 1) +
41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is
clearly divisible by 41.

The incredible formula  n² − 79n + 1601 was discovered, which produces
80 primes for the consecutive values n = 0 to 79. The product of the
coefficients, −79 and 1601, is −126479.

Considering quadratics of the form:

n² + an + b, where |a| < 1000 and |b| < 1000

where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |−4| = 4

Find the product of the coefficients, a and b, for the quadratic
expression that produces the maximum number of primes for consecutive
values of n, starting with n = 0.