The difference between this problem and LTS owns large quantities of apples is more than the data range, please read the statement carefully.

There are $$$n$$$ apples in Hsueh-'s bag, and $$$m$$$ kids around the man. Let's label these children $$$1$$$ to $$$m$$$.

Hsueh- take all of the apples and gave to the first child.

• The first child got $$$n$$$ apples from Hsueh-. Next, he ate an apple and the remaining apples could just be divided into $$$x$$$ piles. He took the $$$x - 1$$$ piles and gave the remaining pile to the second child.
• The second child got $$$\displaystyle \frac{(n - 1)}{x}$$$ apples from the first child. Next, he ate an apple and the remaining apples could just be divided into $$$x$$$ piles. He took the $$$x - 1$$$ piles and gave the remaining pile to the third child.
• $$$\cdots$$$
• The $$$i$$$-th child got some apples from the ($$$i - 1$$$)-th child. Next, he ate an apple and the remaining apples could just be divided into $$$x$$$ piles. He took the $$$x - 1$$$ piles and gave the remaining pile to the ($$$i + 1$$$)-th child.
• $$$\cdots$$$
• The last child got some apples from the Penultimate child, Next, he ate an apple and the remaining apples could just be divided into $$$x$$$ piles. He took $$$x - 1$$$ piles and go away.

It should be noted that the number of apples in a pile left by last child must be a positive integer.

Until the last child go away, Hsueh- wants to know what is the minimum $$$n$$$ that meets all requirements.

Since $$$n$$$ can be very large, output $$$n$$$ modulo $$$998244353$$$. Formally, let $$$M = 998244353$$$. It can be shown that the answer can be expressed as an irreducible fraction $$$\frac{p}{q}$$$, where $$$p$$$ and $$$q$$$ are integers and $$$q \not \equiv 0 \pmod{M}$$$. Output the integer equal to $$$p \cdot q^{-1} \bmod M$$$. In other words, output such an integer $$$x$$$ that $$$0 \le x \lt M$$$ and $$$x \cdot q \equiv p \pmod{M}$$$.