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;
inline void reads(string& x) {
  char kk[((int)2e5 + 123)];
  scanf("%s", kk);
  x = kk;
}
long long ans, n, m, l, r, d;
vector<long long> sum[10];
void solve() {
  scanf("%lld %lld", &n, &m);
  if (m > n || m % 10 == 0) {
    printf("0\n");
    return;
  }
  l = m;
  r = (n / m) * m;
  d = (r - l) / m + 1;
  m = m % 10;
  ans = (d / (long long)sum[m].size()) * sum[m].back();
  d = d % sum[m].size();
  if (d) ans += sum[m][d - 1];
  printf("%lld\n", ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  for (int i = 1, tam; i < 10; i++) {
    tam = 0;
    for (int j = 1; j < 100; j++) {
      if (j % i == 0) {
        sum[i].emplace_back(j % 10);
        if (tam) sum[i][tam] += sum[i][tam - 1];
        tam++;
        if (j % 10 == 0) break;
      }
    }
  }
  int t = 1;
  scanf("%d", &t);
  for (int caso = 1; caso <= t; caso++) {
    solve();
  }
}
