#include <bits/stdc++.h>
int main() {
  long long *n, *m, o, p, y;
  int q, t = 0, sum1, h, sum2;
  scanf("%d", &q);
  m = (long long *)malloc(q * sizeof(long long));
  n = (long long *)malloc(q * sizeof(long long));
  for (int i = 0; i < q; i++) {
    scanf("%lld %lld", n + i, m + i);
  }
  for (int i = 0; i < q; i++) {
    o = 0, sum1 = 0, sum2 = 0;
    t = 0;
    y = m[i] % 10;
    do {
      t++;
      y = (y + (m[i] % 10)) % 10;
      sum1 = sum1 + y;
    } while (y != m[i] % 10);
    o = sum1 * ((n[i] / m[i]) / t);
    h = (n[i] / m[i]) % t;
    for (int f = 0; f < h; f++) {
      sum2 = sum2 + y;
      y = (y + (m[i] % 10)) % 10;
    }
    o = o + sum2;
    printf("%lld\n", o);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  long long *n, *m, o, p, y;
  int q, t = 0, sum1, h, sum2;
  scanf("%d", &q);
  m = (long long *)malloc(q * sizeof(long long));
  n = (long long *)malloc(q * sizeof(long long));
  for (int i = 0; i < q; i++) {
    scanf("%lld %lld", n + i, m + i);
  }
  for (int i = 0; i < q; i++) {
    o = 0, sum1 = 0, sum2 = 0;
    t = 0;
    y = m[i] % 10;
    do {
      t++;
      y = (y + (m[i] % 10)) % 10;
      sum1 = sum1 + y;
    } while (y != m[i] % 10);
    o = sum1 * ((n[i] / m[i]) / t);
    h = (n[i] / m[i]) % t;
    for (int f = 0; f < h; f++) {
      sum2 = sum2 + y;
      y = (y + (m[i] % 10)) % 10;
    }
    o = o + sum2;
    printf("%lld\n", o);
  }
  return 0;
}

```