Living Off the Interest

On New Year's Eve, Juhan made himself a promise: right at the start of the coming year he will open a savings account, into which he will deposit $D$ euros at the end of every month. Juhan's net monthly salary — the amount he actually takes home each month — is $P$ euros. The bank pays interest of $N$ percent per year on the deposit. Juhan now wants to know how many more years he must keep saving before he can start living off the interest the bank pays him.

The bank pays interest once a year, on the last day of the year. If the interest is not enough to live on, Juhan puts the interest paid to him back onto the account that very same day. Juhan can live off the interest only once it exceeds his yearly expenses. For example, if Juhan earns $500$ euros a month and sets aside $200$ euros of it every month, then he needs at least $3600$ euros a year to live on.

On the money that was already in the account at the start of the year, the bank pays $N$ percent interest. On amounts added during the year, the bank pays interest for the number of whole months the money has been kept in the account (money added in December earns no interest, money added in November earns one month of interest, and so on). The interest rate $M$ for one month on deposit is computed as $\left(\left(1 + \frac{N}{100}\right)^{\frac{1}{12}} - 1\right) \cdot 100\%$. The interest rate for $K$ months is computed as $\left(\left(1 + \frac{M}{100}\right)^{K} - 1\right) \cdot 100\%$. The total interest for the year is rounded to the nearest cent and paid out to Juhan.

The only line of input contains three space-separated real numbers $P$, $D$, and $N$ ($P \le 10^6$, $0 < D < P$, $0 < N \le 100$). All numbers are given with a precision of at most $2$ digits after the decimal point.

Print a single integer: the number of years during which Juhan must still go to work.