## Algorithm

B. Smallest number
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers abcd on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.

Input

First line contains four integers separated by space: 0 ≤ a, b, c, d ≤ 1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)

Output

Output one integer number — the minimal result which can be obtained.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

Examples
input
Copy
1 1 1 1
+ + *
output
Copy
3
input
Copy
2 2 2 2
* * +
output
Copy
8
input
Copy
1 2 3 4
* + +
output
Copy
9

## Code Examples

### #1 Code Example with C++ Programming

Code - C++ Programming

#include <cstdio>
#include <vector>
#include <algorithm>

int main(){

const int N = 4;
std::vector<long> num(N); scanf("%ld %ld %ld %ld\n", &num[0], &num[1], &num[2], &num[3]);
std::vector<char> op(N - 1); scanf("%c %c %c\n", &op[0], &op[1], &op[2]);
long long output(1e13);
sort(num.begin(), num.end());

do{
long long current;
if(op[0] == '+'){current = num[0] + num[1];} else{current = num[0] * num[1];}
if(op[1] == '+'){current += num[2];} else{current *= num[2];}
if(op[2] == '+'){current += num[3];} else{current *= num[3];}
if(current < output){output = current;}

long long currentA, currentB;
if(op[0] == '+'){currentA = num[0] + num[1];} else{currentA = num[0] * num[1];}
if(op[1] == '+'){currentB = num[2] + num[3];} else{currentB = num[2] * num[3];}
if(op[2] == '+'){current = currentA + currentB;} else{current = currentA * currentB;}
if(current < output){output = current;}

if(op[0] == '+'){current = num[2] + num[3];} else{current = num[2] * num[3];}
if(op[1] == '+'){current += num[1];} else{current *= num[1];}
if(op[2] == '+'){current += num[0];} else{current *= num[0];}
if(current < output){output = current;}

}while(std::next_permutation(num.begin(), num.end()));

printf("%lld\n", output);

return 0;
}
Copy The Code &

Input

cmd
1 1 1 1
+ + *

Output

cmd
3