Elliptic Curve-based Diffie Hellman: How is the shared key (mnp) calculated? It's a very quick question, but I couldn't find the answer from the sources I have.

I've been coding EC based D-H key exchange, and I'm almost done with it.

So here's what I've understood so far.


Michael and Nikita agree on an Elliptic Curve $E:y^2=x^3+ax+b $, a prime number $p$, and a point $P$ on $E$ over $Z/pZ$.
They choose their secret numbers $m$ (for Michael) and $n$ (for Nikita).
They exchange $mP$ and $nP$, each calculated by adding $P$ to itself $m$ and $n$ times over $E$ in $Z/pZ$. (I understand how the "addition" works)
Now by using the $mP$ or $nP$ they got from each other and the secret number they chose ($n$ and $m$) to calculate the shared key $mnP$.


So my question is, how do you calculate $mnP$? No source I have specified what exactly it is, so I'm not sure if it is $(m \times n)P$ or $P$ added $n$ times to $mP$.

Thanks in advance!
So my question is, how do you calculate $mnP$?


For Michael, he takes the value $nP$ he received from Nikita, and adds it to itself $m$ times; that is, it's $\underbrace{nP + nP + \dots + nP}_\text{$m$ times}$, that is the value $m(nP)$.

And, of course Nikita takes the value $mP$ he received from Michael, and adds it to itself $n$ times; that is, it's $\underbrace{mP + mP + \dots + mP}_\text{$n$ times}$, that is the value $n(mP)$.

And, because we perform these operations in a mathematical group (that is, associativity holds), we have $m(nP) = n(mP) = mnP$, so they come up with the same value.