CodeForces - 194A Exams

A. Exams

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

4 8

output

4

input

4 10

output

2

input

1 3

output

0

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.

k一定大于等于2*n，当k>=3*n时，就不能考2分，当k==2*n时，全部都能考2分。主要考虑k>2*n&&k<3*n的情况。首先在这个范围内肯定是需要2的，所以我们先让每一门都考2分，看最后差几分，把差的分给几门课，剩下的就是考2分的课程，具体怎么分我们不用管。因为k此时的取值范围，所以差的分数不可能超过n。 所以此时考2分的最小数目为n-(k-2*n)。

#include
int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(3*n<=k)
            printf("%d\n",0);
        else
            printf("%d\n",n-(k-2*n));
    }

}