#include <bits/stdc++.h>
using namespace std;
template <class T>
void show(const vector<T> &a) {
  for (T x : a) cout << x << " ";
  cout << '\n';
}
const long long N = 3e5 + 50, oo = 1e18 + 100, mod = 1e9 + 7;
const long double eps = 1e-9, PI = 2 * acos(0.0);
vector<long long> g[N];
vector<long long> visit(N, 0);
long long n, m, k;
long long cnt = 0;
signed main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long b, p, f, h, c;
    cin >> b >> p >> f >> h >> c;
    cnt = 0;
    b /= 2;
    if (h > c) {
      long long x = min(b, p);
      b -= x;
      cnt += x * h;
      x = min(b, f);
      cnt += x * c;
    } else {
      long long x = min(b, f);
      b -= x;
      cnt += x * c;
      x = min(b, p);
      cnt += x * h;
    }
    cout << cnt << '\n';
  }
}

### Prompt

Generate 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 <class T>
void show(const vector<T> &a) {
  for (T x : a) cout << x << " ";
  cout << '\n';
}
const long long N = 3e5 + 50, oo = 1e18 + 100, mod = 1e9 + 7;
const long double eps = 1e-9, PI = 2 * acos(0.0);
vector<long long> g[N];
vector<long long> visit(N, 0);
long long n, m, k;
long long cnt = 0;
signed main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long b, p, f, h, c;
    cin >> b >> p >> f >> h >> c;
    cnt = 0;
    b /= 2;
    if (h > c) {
      long long x = min(b, p);
      b -= x;
      cnt += x * h;
      x = min(b, f);
      cnt += x * c;
    } else {
      long long x = min(b, f);
      b -= x;
      cnt += x * c;
      x = min(b, p);
      cnt += x * h;
    }
    cout << cnt << '\n';
  }
}

```