Why do the weights of a generic Newton-Cotes quadrature formula show symmetry w.r.t. the midpoint of the integration interval? I understand that, given $f:[a,b]\mapsto\mathbb{R}$ and being $n>0$ the number of nodes of interpolation, then the definite integral of Lagrange's basis generators $\ell_j(x)$ defines the weights $w_j$ of the quadrature formula; of course since $\ell_j(x)$ only depends on the nodes chosen and not on $f$ then the form that $w_j$ takes solely depends on the aforementioned nodes.
However I do not understand where the symmetry $w_j=w_{n+1-j}$ arises since $\ell_j(x)\neq\ell_{n+1-j}$.
Also is this valid only if $n$ is odd (i.e. if the midpoint does coincide with a node) or when $n$ is even also?
PS: The nodes are all equidistant from one another; I understand that the definition of the Newton-Cotes formula implicitly defines a uniform partition of $[a,b]$ in $n-1$ intervals of width $h=x_{j+1}-x_j\,,\;\forall j=1,...,n-1$.