Game

Rang, who is good at math, learned a number game from her butler today. The game starts with a positive integer $N$ as the initial value. Let $A$ be the sum of the digits of $N$, and let $B$ be the product of the digits of $N$. Then $f(N)$ is the number formed by writing $A$ and $B$ one after another. For example, if $N = 12352$, the digit sum is $1 + 2 + 3 + 5 + 2 = 13$ and the digit product is $1 \times 2 \times 3 \times 5 \times 2 = 60$, so $f(N)$ is $1360$. Applying the same operation to $1360$ gives $1 + 3 + 6 + 0 = 10$ and $1 \times 3 \times 6 \times 0 = 0$, so $f(1360) = 100$. Applying it to $100$ gives $f(100) = 10$, and applying it to $10$ gives $f(10) = 10$. In this game, given $N$, the important question is whether repeatedly computing $f(N)$, $f(f(N))$, $f(f(f(N)))$, $\cdots$ eventually reaches some $x$ with $f(x) = x$.

If repeatedly applying the operation to $N$ reaches some $x$ with $x = f(x)$, let $g(N) = 1$. If it never does, let $g(N) = 0$. However, if any of $N, f(N), f(f(N)), \cdots$ is greater than $100\,000$, the computation is hard, so $g(N) = -1$. For example, when $N = 1$, we get $f(1) = 11$, $f(11) = 21$, $f(21) = 32$, $f(32) = 56$, $f(56) = 1130$, $f(1130) = 50$, and $f(50) = 50$. Since $f(x) = x$ holds at $50$, $g(1) = 1$.

Given positive integers $L$ and $R$ ($L \le R$), write a program that computes $g(L) + g(L + 1) + \cdots + g(R)$.

The first line contains positive integers $L$ and $R$, separated by a space. ($1 \le L \le R \le 100\,000$)

Print $g(L) + g(L + 1) + \cdots + g(R)$.