Euler discovered the remarkable quadratic formula:

$n^2+n+41$

It turns out that the formula will produce $40$ primes for the consecutive integer values $0\le n\le 39$. However, when $n=40,{40}^2+40+41=40(40+1)+41$ is divisible by $41$, and certainly when $n=41,{41}^2+41+41$ is clearly divisible by $41$.

The incredible formula $n^2-79n+1601$ was discovered, which produces $80$ primes for the consecutive values $0\le n\le 79$. The product of the coefficients, $-79$ and $1601$, is $-126479$.

Considering quadratics of the form:

$n^2+an+b$, where $|a|<1000$ and $|b|\le 1000$

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 $n=0$.