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>
using namespace std;
int a[10][200];
int sum[10];
int pnum[10] = {0};
void init() {
  for (int i = 1; i <= 9; i++) {
    pnum[i] = 1;
    a[i][pnum[i]++] = i;
    sum[i] = i;
    while (a[i][1] != (i * pnum[i] % 10)) {
      a[i][pnum[i]] = (i * (pnum[i]) % 10);
      sum[i] += a[i][pnum[i]++];
    }
    pnum[i]--;
  }
}
int main() {
  init();
  int t;
  cin >> t;
  long long n, m, res;
  while (t--) {
    scanf("%I64d %I64d", &n, &m);
    long long num = n / m;
    res = 0;
    if (m % 10 == 0) {
      printf("0\n");
      continue;
    }
    res = (num / (pnum[m % 10])) * sum[m % 10];
    long long temp = num % (pnum[m % 10]);
    for (int i = 1; i <= temp; i++) {
      res += a[m % 10][i];
    }
    printf("%I64d\n", res);
  }
}
