Crossing the River

Doctor Nefario is leaving on a trip after a long stretch of work in his lab. On the way he has to cross a river. The river holds $N$ rocks laid out on a straight line to the far bank, and he can use them as stepping stones. The width of the river, meaning the whole distance he has to cover, is $L$.

The doctor's scooter hovers but cannot jump far, so a longer jump needs a rocket booster. A booster costs more when it is stronger. A booster that jumps up to distance $R$ costs $R^2$. The booster works any number of times, but every jump costs another $C$. For example, a booster of range $10$ used for five jumps costs $10^2 + 5 \times C = 100 + 5C$ in total.

He starts on the near bank at distance $0$ and finishes on the far bank at distance $L$. One jump moves him forward by at most $R$, and he must land on a rock or on the far bank.

Given the width $L$, the cost $C$ of one jump, and the positions of the $N$ rocks, find the minimum cost $M$ of crossing the river, the number of jumps $J$ it takes, and the booster range $R$.

In the figure the doctor crosses a river of width $6$. The rocks sit at distances $1$, $2$, $3$, and $5$. A booster of range $3$ carries him across in two jumps.

The limits are $1 \le L \le 10^9$, $0 \le C \le 10^6$, and $0 \le N < 1000$. A rock position is an integer greater than $0$ and less than $L$. The range $R$ is chosen as a positive integer. The values grow large enough to need 64-bit integers, and scanning every possible range does not fit in the time limit.

The input holds several test cases. The first line of each test case has the integers $L$, $C$, and $N$: the width of the river, the cost of one jump, and the number of rocks. Each of the next $N$ lines holds the position of one rock. The rocks are not always listed in order of position. The last line holds a single $0$, and the input ends on that line.

For each test case print the minimum cost of crossing the river on one line, in this format.

Minimum cost M achieved with J jumps of range R

$M$ is the minimum cost, $R$ is the range that reaches it, and $J$ is the smallest number of jumps needed to cross with range $R$. When several ranges reach the minimum cost, print the smallest of them as $R$.