Taming the Herd

Early one morning, Farmer John woke up to the sound of splintering wood. The cows were breaking out of the barn again.

Farmer John had had enough of the morning breakouts, so he nailed a counter to the barn wall. The counter shows how many days have passed since the last breakout. If a breakout happens in the morning, the counter reads $0$ that day. If the most recent breakout was $3$ days ago, the counter reads $3$. Farmer John wrote the counter down every single day.

The year is over and Farmer John wants to settle accounts with the cows, but something about his log looks wrong. He suspects the cows tampered with it. The one thing he is sure of is that he started the log on a day when a breakout happened.

For each possible number of breakouts since the log began, find the smallest number of log entries that must have been tampered with.

The first line contains one integer $N$ ($1 \leq N \leq 100$), the number of days Farmer John recorded.

The second line contains $N$ space separated integers. If the cows did not tamper with the entry for day $i$, the counter on that day read $a_i$ ($0 \leq a_i \leq 100$).

Print $N$ lines. The $i$th line contains the minimum, taken over every breakout sequence with exactly $i$ breakouts, of the number of log entries that disagree with that sequence.

Take the six day log 1 1 2 0 0 1.

With one breakout the correct log is 0 1 2 3 4 5, which disagrees with the given log in 4 entries. With two breakouts the correct log can be 0 1 2 3 0 1, which disagrees in 2 entries, and the breakouts fall on day 1 and day 5. With three breakouts the correct log can be 0 1 2 0 0 1, which disagrees in 1 entry, and the breakouts fall on day 1, day 4 and day 5. Larger counts work the same way.