#include <bits/stdc++.h>
using namespace std;
template <typename T, typename T1>
ostream &operator<<(ostream &out, pair<T, T1> obj) {
  out << "(" << obj.first << ", " << obj.second << ")";
  return out;
}
template <typename T, typename T1>
ostream &operator<<(ostream &out, map<T, T1> cont) {
  typename map<T, T1>::const_iterator itr = cont.begin();
  typename map<T, T1>::const_iterator ends = cont.end();
  for (; itr != ends; ++itr) {
    out << *itr << " ";
  }
  out << endl;
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, set<T> cont) {
  typename set<T>::const_iterator itr = cont.begin();
  typename set<T>::const_iterator ends = cont.end();
  for (; itr != ends; ++itr) {
    out << *itr << " ";
  }
  out << endl;
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, multiset<T> cont) {
  typename multiset<T>::const_iterator itr = cont.begin();
  typename multiset<T>::const_iterator ends = cont.end();
  for (; itr != ends; ++itr) {
    out << *itr << " ";
  }
  out << endl;
  return out;
}
template <typename T,
          template <typename ELEM, typename ALLOC = allocator<ELEM>> class CONT>
ostream &operator<<(ostream &out, CONT<T> cont) {
  typename CONT<T>::const_iterator itr = cont.begin();
  typename CONT<T>::const_iterator ends = cont.end();
  for (; itr != ends; ++itr) {
    out << *itr << " ";
  }
  out << endl;
  return out;
}
template <typename T, unsigned int N, typename CTy, typename CTr>
typename enable_if<!is_same<T, char>::value, basic_ostream<CTy, CTr> &>::type
operator<<(basic_ostream<CTy, CTr> &out, const T (&arr)[N]) {
  for (auto i = 0; i < N; ++i) {
    out << arr[i] << " ";
  }
  out << endl;
  return out;
}
template <typename T>
T GCD(T a, T b) {
  T min_v = min(a, b);
  T max_v = max(a, b);
  while (min_v) {
    T temp = max_v % min_v;
    max_v = min_v;
    min_v = temp;
  }
  return max_v;
}
template <typename T>
T LCM(T a, T b) {
  return (a * b) / gcd(a, b);
}
template <typename T>
T fastExpPow(T base, T exp, T mod) {
  T res = 1;
  while (exp) {
    if (exp & 1) {
      res *= base;
      res %= mod;
    }
    exp >>= 1;
    base *= base;
    base %= mod;
  }
  return res % mod;
}
int T, B, P, F, H, C;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d%d%d", &B, &P, &F, &H, &C);
    int profit = 0;
    if (H >= C) {
      int many = min(B / 2, P);
      B -= 2 * many;
      profit += many * H;
      many = min(B / 2, F);
      profit += many * C;
    } else {
      int many = min(B / 2, F);
      B -= 2 * many;
      profit += many * C;
      many = min(B / 2, P);
      profit += many * H;
    }
    printf("%d\n", profit);
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

There are two types of burgers in your restaurant — hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet. 

You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.

You have to answer t independent queries.

Input

The first line contains one integer t (1 ≤ t ≤ 100) – the number of queries.

The first line of each query contains three integers b, p and f (1 ≤ b, ~p, ~f ≤ 100) — the number of buns, beef patties and chicken cutlets in your restaurant.

The second line of each query contains two integers h and c (1 ≤ h, ~c ≤ 100) — the hamburger and chicken burger prices in your restaurant.

Output

For each query print one integer — the maximum profit you can achieve.

Example

Input


3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100


Output


40
34
0

Note

In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 ⋅ 5 + 3 ⋅ 10 = 40.

In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 ⋅ 10 + 2 ⋅ 12 = 34.

In third query you can not create any type of burgers because because you have only one bun. So your income is zero.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename T1>
ostream &operator<<(ostream &out, pair<T, T1> obj) {
  out << "(" << obj.first << ", " << obj.second << ")";
  return out;
}
template <typename T, typename T1>
ostream &operator<<(ostream &out, map<T, T1> cont) {
  typename map<T, T1>::const_iterator itr = cont.begin();
  typename map<T, T1>::const_iterator ends = cont.end();
  for (; itr != ends; ++itr) {
    out << *itr << " ";
  }
  out << endl;
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, set<T> cont) {
  typename set<T>::const_iterator itr = cont.begin();
  typename set<T>::const_iterator ends = cont.end();
  for (; itr != ends; ++itr) {
    out << *itr << " ";
  }
  out << endl;
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, multiset<T> cont) {
  typename multiset<T>::const_iterator itr = cont.begin();
  typename multiset<T>::const_iterator ends = cont.end();
  for (; itr != ends; ++itr) {
    out << *itr << " ";
  }
  out << endl;
  return out;
}
template <typename T,
          template <typename ELEM, typename ALLOC = allocator<ELEM>> class CONT>
ostream &operator<<(ostream &out, CONT<T> cont) {
  typename CONT<T>::const_iterator itr = cont.begin();
  typename CONT<T>::const_iterator ends = cont.end();
  for (; itr != ends; ++itr) {
    out << *itr << " ";
  }
  out << endl;
  return out;
}
template <typename T, unsigned int N, typename CTy, typename CTr>
typename enable_if<!is_same<T, char>::value, basic_ostream<CTy, CTr> &>::type
operator<<(basic_ostream<CTy, CTr> &out, const T (&arr)[N]) {
  for (auto i = 0; i < N; ++i) {
    out << arr[i] << " ";
  }
  out << endl;
  return out;
}
template <typename T>
T GCD(T a, T b) {
  T min_v = min(a, b);
  T max_v = max(a, b);
  while (min_v) {
    T temp = max_v % min_v;
    max_v = min_v;
    min_v = temp;
  }
  return max_v;
}
template <typename T>
T LCM(T a, T b) {
  return (a * b) / gcd(a, b);
}
template <typename T>
T fastExpPow(T base, T exp, T mod) {
  T res = 1;
  while (exp) {
    if (exp & 1) {
      res *= base;
      res %= mod;
    }
    exp >>= 1;
    base *= base;
    base %= mod;
  }
  return res % mod;
}
int T, B, P, F, H, C;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d%d%d", &B, &P, &F, &H, &C);
    int profit = 0;
    if (H >= C) {
      int many = min(B / 2, P);
      B -= 2 * many;
      profit += many * H;
      many = min(B / 2, F);
      profit += many * C;
    } else {
      int many = min(B / 2, F);
      B -= 2 * many;
      profit += many * C;
      many = min(B / 2, P);
      profit += many * H;
    }
    printf("%d\n", profit);
  }
  return 0;
}

```