# Why does this definition of the 3-PARTITION problem imply that every set contains exactly 3 elements?

I have the following definition of the 3-PARTITION problem taken from this paper: https://www.sciencedirect.com/science/article/pii/0166218X93900853

Given $$3m$$ positive integers $$a_1, a_2,...,a_{3m}$$ and a positive integral bound $$B$$, the integers $$a_i, i = 1, \dots , 3m$$ satisfy the the conditions:

$$\frac{B}{4} < a_i < \frac{B}{2}$$

and

$$\sum\limits_{i=1}^{3m}a_i = mB$$

Can $$I = \{1,2,\dots,3m\}$$ be partitioned into $$m$$ disjoint sets $$I_1,I_2,\dots,I_m$$ such that for $$j, \dots, m$$.

$$\sum_{i \in I_j} a_i = B$$

So this all makes sense- basic 3 partition problem. However, the next line says this:

"Notice that the conditions $$\frac{B}{4} < a_i < \frac{B}{2}$$ imply that every set $$I_j$$ with $$\sum_{i\in I_J}a_i = B$$ must contain exactly three elements."

I must be missing something because I do not understand how this implies that every set $$I_j$$ will contain exactly 3 elements?

If $$I_j$$ contains two or fewer elements, then $$\sum_{i \in I_j} a_i<2\dfrac{B}{2}=B$$. If $$I_j$$ contains four or more elements, then $$\sum_{i \in I_j} a_i>4\dfrac{B}{4}=B$$. So in neither case can $$\sum_{i \in I_j}$$ be equal to $$B$$.