Codeforces Round #642 (Div. 3) A. Most Unstable Array

You are given two integers nn and mm . You have to construct the array aa of length nn consisting of non-negative integers (i.e. integers greater than or equal to zero) such that the sum of elements of this array is exactly mm and the value ∑i=1n−1|ai−ai+1|∑i=1n−1|ai−ai+1| is the maximum possible. Recall that |x||x| is the absolute value of xx .

In other words, you have to maximize the sum of absolute differences between adjacent (consecutive) elements. For example, if the array a=[1,3,2,5,5,0]a=[1,3,2,5,5,0] then the value above for this array is |1−3|+|3−2|+|2−5|+|5−5|+|5−0|=2+1+3+0+5=11|1−3|+|3−2|+|2−5|+|5−5|+|5−0|=2+1+3+0+5=11 . Note that this example doesn't show the optimal answer but it shows how the required value for some array is calculated.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1041≤t≤104 ) — the number of test cases. Then tt test cases follow.

The only line of the test case contains two integers nn and mm (1≤n,m≤1091≤n,m≤109 ) — the length of the array and its sum correspondingly.

Output

For each test case, print the answer — the maximum possible value of ∑i=1n−1|ai−ai+1|∑i=1n−1|ai−ai+1| for the array aa consisting of nn non-negative integers with the sum mm .

Example

Input

Copy

5
1 100
2 2
5 5
2 1000000000
1000000000 1000000000

Output

Copy

0
2
10
1000000000
2000000000
挺有趣的思维水题。
可以注意到，n=1时答案为0，n=2时答案为m，n=3时整个差分之和不超过2*m，即答案为2*m.最大的情况就是0 a1 0 a2 0 a3...这么构造（a1+a2+a3..=m）或者0 0 0 m 0 0 0（m一定要放在中间)

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        if(n==1) cout<<0<<endl;
        else if(n==2)cout<<m<<endl;
        else cout<<2*m<<endl;
    }
}