Polycarp is reading a book consisting of n pages numbered from 1 to n. Every time he finishes the page with the number divisible by m, he writes down the last digit of this page number. For example, if n=15 and m=5, pages divisible by m are 5, 10, 15. Their last digits are 5, 0, 5 correspondingly, their sum is 10.

Your task is to calculate the sum of all digits Polycarp has written down.

You have to answer q independent queries.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 1000) — the number of queries.

The following q lines contain queries, one per line. Each query is given as two integers n and m (1 ≤ n, m ≤ 10^{16}) — the number of pages in the book and required divisor, respectively.

Output

For each query print the answer for it — the sum of digits written down by Polycarp.

Example

Input


7
1 1
10 1
100 3
1024 14
998244353 1337
123 144
1234312817382646 13


Output


1
45
153
294
3359835
0
427262129093995
#include <bits/stdc++.h>
int main() {
  long long int n, m, i, ans = 0, cish, h, sd = 0, c;
  int q, j;
  scanf("%d", &q);
  int a[10][10] = {{0},
                   {1, 2, 3, 4, 5, 6, 7, 8, 9, 0},
                   {2, 4, 6, 8, 0},
                   {3, 6, 9, 2, 5, 8, 1, 4, 7, 0},
                   {4, 8, 2, 6, 0},
                   {5, 0},
                   {6, 2, 8, 4, 0},
                   {7, 4, 1, 8, 5, 2, 9, 6, 3, 0},
                   {8, 6, 4, 2, 0},
                   {9, 8, 7, 6, 5, 4, 3, 2, 1, 0}};
  for (j = 0; j < q; j++) {
    scanf("%lld%lld", &n, &m);
    c = m % 10;
    switch (c) {
      case 1:
      case 3:
      case 7:
      case 9:
        cish = 10;
        h = 45;
        break;
      case 2:
      case 4:
      case 6:
      case 8:
        cish = 5;
        h = 20;
        break;
      case 5:
        cish = 2;
        h = 5;
        break;
      case 0:
        cish = 1;
        h = 0;
        break;
    }
    int y;
    for (y = 0; y < (n / m) % cish; y++) sd += a[c][y];
    ans = (n / m) / cish * h + sd;
    printf("%lld\n", ans);
    ans = 0;
    sd = 0;
  }
  return 0;
}
