One day the Codeforces round author sat exams. He had $*n*$ exams and he needed to get an integer from $2$ to $5$ for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark $2$.

The author would need to spend too much time and effort to make the sum of his marks strictly more than $*k*$. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than $*k*$, the author's mum won't be pleased at all.

The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.

Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all $*n*$ exams equal exactly $*k*$.

Input

The single input line contains space-separated integers $*n*$ and $*k*$ ($1 ≤ *n* ≤ 50$, $1 ≤ *k* ≤ 250$) — the number of exams and the required sum of marks.

It is guaranteed that there exists a way to pass $*n*$ exams in the way that makes the sum of marks equal exactly $*k*$.

Output

Print the single number — the minimum number of exams that the author will get a $2$ for, considering that the sum of marks for all exams must equal $*k*$.

Note

In the first sample the author has to get a $2$ for all his exams.

In the second sample he should get a $3$ for two exams and a $2$ for two more.

In the third sample he should get a $3$ for one exam.