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 main() {
  long long q;
  cin >> q;
  long long dp[11][11];
  for (long long i = 1; i <= 9; i++) {
    if (i == 1 || i == 3 || i == 7 || i == 9) {
      dp[i][1] = i;
      for (long long j = 2; j <= 10; j++) dp[i][j] = (dp[i][j - 1] + i) % 10;
    } else if (i != 5) {
      dp[i][1] = i;
      for (long long j = 2; j <= 5; j++) dp[i][j] = (dp[i][j - 1] + i) % 10;
    } else {
      dp[i][1] = 5;
      dp[i][2] = 0;
    }
  }
  for (long long i = 1; i <= 9; i++) {
    if (i == 1 || i == 3 || i == 7 || i == 9) {
      for (long long j = 2; j <= 10; j++) dp[i][j] += dp[i][j - 1];
    } else if (i != 5) {
      for (long long j = 2; j <= 5; j++) dp[i][j] += dp[i][j - 1];
    } else {
      dp[i][2] = 5;
    }
  }
  while (q--) {
    long long n, m;
    cin >> n >> m;
    long long e = 5;
    long long t = m % 10;
    if (t == 0)
      cout << 0 << endl;
    else {
      if (t == 1 || t == 3 || t == 7 || t == 9)
        e = 10;
      else if (t == 5)
        e = 2;
      long long y = n / m;
      y = y / e;
      long long z = (n / m) % e;
      long long ans = dp[t][e] * y;
      if (z) ans += dp[t][z];
      cout << ans << endl;
    }
  }
}
