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;
bool c[11];
int go(long long x) {
  int cnt = 0;
  while (x >= 10) {
    x /= 10;
    cnt += 1;
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q;
  cin >> q;
  while (q--) {
    long long n, m;
    cin >> n >> m;
    long long tmp = m % (long long)10;
    memset(c, false, sizeof(c));
    long long j = 2;
    long long sum = 0;
    long long cnt = 0;
    for (int i = 1; i <= 100; i++) {
      long long C = (tmp * i) % 10;
      if (!c[C]) {
        c[C] = true;
        sum += C;
        cnt += 1;
      }
    }
    long long l = 0;
    long long r = 10000000000000000;
    while (l <= r) {
      long long mid = (l + r) / 2;
      long long Cnt = mid * m;
      int Tmp = go(mid) + go(m);
      if (Tmp >= 16 || Cnt < 0 || Cnt > n) {
        r = mid - 1;
        continue;
      } else {
        l = mid + 1;
      }
    }
    long long ans = sum * (r / cnt);
    for (int i = 0; i < (r % cnt); i++) {
      ans += ((long long)(i + 1) * tmp) % (long long)10;
    }
    cout << ans << '\n';
  }
}
