Water Buying

Posted 2021-02-11 bravewtz

Water Buying

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp wants to cook a soup. To do it, he needs to buy exactly nn liters of water.

There are only two types of water bottles in the nearby shop — 11 -liter bottles and 22 -liter bottles. There are infinitely many bottles of these two types in the shop.

The bottle of the first type costs aa burles and the bottle of the second type costs bb burles correspondingly.

Polycarp wants to spend as few money as possible. Your task is to find the minimum amount of money (in burles) Polycarp needs to buy exactly nn liters of water in the nearby shop if the bottle of the first type costs aa burles and the bottle of the second type costs bb burles.

You also have to answer qq independent queries.

Input

The first line of the input contains one integer qq (1≤q≤5001≤q≤500 ) — the number of queries.

The next nn lines contain queries. The ii -th query is given as three space-separated integers nini , aiai and bibi (1≤ni≤1012,1≤ai,bi≤10001≤ni≤1012,1≤ai,bi≤1000 ) — how many liters Polycarp needs in the ii -th query, the cost (in burles) of the bottle of the first type in the ii -th query and the cost (in burles) of the bottle of the second type in the ii -th query, respectively.

Output

Print qq integers. The ii -th integer should be equal to the minimum amount of money (in burles) Polycarp needs to buy exactly nini liters of water in the nearby shop if the bottle of the first type costs aiai burles and the bottle of the second type costs bibi burles.

Input

4
10 1 3
7 3 2
1 1000 1
1000000000000 42 88
Output

10
9
1000
42000000000000

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 #define ll long long 
 5 int main(int argc, char const *argv[])
 6 {
 7     int n;
 8     cin>>n;
 9     ll k,a,b;
10     for( int i=0; i<n; i++ ){
11         cin>>k>>a>>b;
12         if(a*2<=b) cout<<k*a<<endl;
13         else{
14             if(k%2==0) cout<<k/2*b<<endl;
15             else cout<<k/2*b+a<<endl;
16         }
17     }
18     return 0;
19 }