Problem:
Score : 400 points

Problem Statement

You start with the number 0 and you want to reach the number N.

You can change the number, paying a certain amount of coins, with the following operations:

Multiply the number by 2, paying A coins.

Multiply the number by 3, paying B coins.

Multiply the number by 5, paying C coins.

Increase or decrease the number by 1, paying D coins.

You can perform these operations in arbitrary order and an arbitrary number of times.

What is the minimum number of coins you need to reach N?

You have to solve T testcases.

Constraints

1 \le T \le 10

1 \le N \le 10^{18}

1 \le A, B, C, D \le 10^9

All numbers N, A, B, C, D are integers.

Input

The input is given from Standard Input. The first line of the input is

T

Then, T lines follow describing the T testcases.
Each of the T lines has the format

N A B C D

Output

For each testcase, print the answer on Standard Output followed by a newline.

Sample Input 1

5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321

Sample Output 1

20
19
26
3821859835
23441258666

For the first testcase, a sequence of moves that achieves the minimum cost of 20 is:

Initially x = 0.

Pay 8 to increase by 1 (x = 1).

Pay 1 to multiply by 2 (x = 2).

Pay 1 to multiply by 2 (x = 4).

Pay 2 to multiply by 3 (x = 12).

Pay 8 to decrease by 1 (x = 11).

For the second testcase, a sequence of moves that achieves the minimum cost of 19 is:

Initially x = 0.

Pay 8 to increase by 1 (x = 1).

Pay 1 to multiply by 2 (x = 2).

Pay 2 to multiply by 5 (x = 10).

Pay 8 to increase by 1 (x = 11).

Solution in PHP (7.4.4) (judge verdict: Wrong Answer)

<?php
// 入力：ループ回数取得
$t = fgets(STDIN);
for ($j=0; $j<$t; $j++){
  // 入力：テストケース取得
  fscanf(STDIN, "%d %d %d %d %d", $n,$a,$b,$c,$d);

  if ($n == 0){
    echo 0;
  }
  elseif ($n == 1){
    echo $d;
  }
  else{
    $amount = -1;

    for ($i=-1; $i<2; $i++){
      $x = $n+$i;
      $anum = 0;
      $bnum = 0;
      $cnum = 0;

      while ($x%2 == 0) {
        $anum +=$a;
        $x = $x/2;
      }
      while ($x%3 == 0) {
        $bnum +=$b;
        $x = $x/3;
      }
    
      while ($x%5 == 0) {
        $cnum +=$c;
        $x = $x/5;
      }

      if ($x == 1){
        $t_amount = $anum + $bnum + $cnum + abs($i)*$d;
      }
      else {
        $t_amount = $anum + $bnum + $cnum + $x*$d;
      }

      echo "t". $anum . $bnum . $cnum. $t_amount . "\n";
      if($amount < 0 || $t_amount < $amount) $amount = $t_amount;
      echo "a".$amount . "\n";
    }

    // 初回の操作Ｄ＋と以降の操作の合計
    echo $d + $amount . "\n";
  }
}
?>