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 num[15][15];
int sum[15][15];
int beishu[15];
void pre() {
  int i, j, k;
  for (i = 1; i <= 9; i++) {
    for (j = 1; j <= 10; j++) {
      k = (j * i) % 10;
      num[i][j] = k;
      sum[i][j] = sum[i][j - 1] + num[i][j];
      if (!k) {
        beishu[i] = j;
        break;
      }
    }
  }
}
int main() {
  int q;
  long long n, m, k, b, y, now, b1;
  scanf("%d", &q);
  pre();
  while (q--) {
    long long ans = 0;
    scanf("%I64d %I64d", &n, &m);
    if (m > n || ((m % 10) == 0)) {
      printf("0\n");
      continue;
    }
    if (n == m) {
      printf("%d\n", m % 10);
      continue;
    }
    k = m % 10;
    b = n / m;
    b1 = b / beishu[k];
    y = b % beishu[k];
    ans += b1 * (long long)sum[k][beishu[k]];
    ans += (long long)sum[k][y];
    printf("%I64d\n", ans);
  }
}
