--description--

An integer partition of a number $n$ is a way of writing $n$ as a sum of positive integers.

Partitions that differ only in the order of their summands are considered the same. A partition of $n$ into distinct parts is a partition of $n$ in which every part occurs at most once.

The partitions of 5 into distinct parts are:

5, 4 + 1 and 3 + 2.

Let $f(n)$ be the maximum product of the parts of any such partition of $n$ into distinct parts and let $m(n)$ be the number of elements of any such partition of $n$ with that product.

So $f(5) = 6$ and $m(5) = 2$.

For $n = 10$ the partition with the largest product is $10 = 2 + 3 + 5$, which gives $f(10) = 30$ and $m(10) = 3$. And their product, $f(10) \times m(10) = 30 \times 3 = 90$

It can be verified that $\sum f(n) \times m(n)$ for $1 ≤ n ≤ 100 = 1\,683\,550\,844\,462$.

Find $\sum f(n) \times m(n)$ for $1 ≤ n ≤ {10}^{14}$. Give your answer modulo $982\,451\,653$, the 50 millionth prime.

--hints--

maximumIntegerPartitionProduct() should return 334420941.

assert.strictEqual(maximumIntegerPartitionProduct(), 334420941);

--seed--

--seed-contents--

function maximumIntegerPartitionProduct() {

  return true;
}

maximumIntegerPartitionProduct();

--solutions--

// solution required