np complete – A variant of partition problem

I am thinking about the following variant version of partition problem:

Given $n$ positive integer numbers $a_1, ldots, a_n$ which sum to a even number $S= sum_{i=1}^n a_i$. The decision problem of the existence of a subset $A$ such that $sum_{i in A}a_i = S/2 + 1$ is NP-hard?

Here I’m essentially partitioning the ground set ${1,ldots, n}$ into two sets whose summation differ by $textbf{exactly}$ 2. If this difference is $textbf{at most}$ 2 instead, then we can obtain a simple reduction from partition problem by multiplying every numbers by 3.