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 long long gcd(long long a, long long b) {
  while (b != 0) {
    long long c = a % b;
    a = b;
    b = c;
  }
  return a < 0 ? -a : a;
}
inline long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
inline long long lowbit(long long x) { return x & (-x); }
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
inline long long rd() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const double eps = 1e-8;
const int M = 4e6 + 10;
const int N = 1e6 + 10;
bool vis[11];
int main() {
  int t = rd();
  while (t--) {
    long long n = rd(), m = rd();
    memset(vis, 0, sizeof vis);
    long long ans = 0;
    long long cnt = 0, sum = 0;
    long long tmp = n;
    for (long long i = m; i <= n; i += m) {
      long long x = i % 10;
      if (vis[x]) {
        tmp = i - m;
        break;
      }
      vis[x] = 1;
      sum += x;
    }
    ans += (n / tmp) * sum;
    for (long long i = (n / tmp) * tmp + m; i <= n; i += m) {
      ans += i % 10;
    }
    cout << ans << endl;
  }
}
