How can I change the default grouping on an operator without a built-in meaning?

I've created my own infix operator by defining LeftArrow.

SetAttributes[LeftArrow, {NumericFunction, OneIdentity}]
ex1_ \[LeftArrow] ex2_ := ex1 /. Rule[ex2[[1]], ex2[[2]]]

(I'm using the Esc<-Esc form of LeftArrow in the second line, but I'm not sure how to get it to cut and paste...)

When I use it, I need to string together applications like so:

eqIld2 = (((((((eqIld \[LeftArrow] eqVrx) \[LeftArrow] eqVct) \[LeftArrow] eqIcr) \[LeftArrow] eqVct) \[LeftArrow] eqIcl) \[LeftArrow] eqVtx) \[LeftArrow] eqIin)

I'd like to be able to avoid all the parentheses, and to get the same result for the same input with the parentheses removed.

I don't see the default grouping for LeftArrow documented anywhere. I've tried playing with various Attributes, but I can't find one that does what I want. It looks like there's an InfixNotation that accepts options, but they aren't documented.

What's the trick?

How can I change the default grouping on an operator without a built-in meaning?

I've created my own infix operator by defining LeftArrow.

SetAttributes[LeftArrow, {NumericFunction, OneIdentity}]
ex1_ ← ex2_ := ex1 /. Rule[ex2[[1]], ex2[[2]]]

(I'm using the Esc<-Esc form of LeftArrow in the second line)

When I use it, I need to string together applications like so:

eqIld2 = (((((((eqIld ← eqVrx) ← eqVct) ← eqIcr) ← eqVct) ← eqIcl) ← eqVtx) ← eqIin)

I'd like to be able to avoid all the parentheses, and to get the same result for the same input with the parentheses removed.

I don't see the default grouping for LeftArrow documented anywhere. I've tried playing with various Attributes, but I can't find one that does what I want. It looks like there's an InfixNotation that accepts options, but they aren't documented.

What's the trick?