#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, m;
  cin >> n >> m;
  if (m > n or m % 10 == 0) {
    cout << 0 << endl;
    return;
  }
  long long a[12];
  for (int i = 1; i <= 10; i++) {
    a[i] = (i * m) % 10;
  }
  if (m % 2 == 0) {
    long long x = n / m;
    long long ans = (x / 5) * (20ll);
    for (int i = 1; i <= x % 5; i++) ans += a[i];
    cout << ans << endl;
    return;
  }
  if (m % 10 == 5) {
    long long x = n / m;
    long long ans = (x / 2) * (5ll);
    for (int i = 1; i <= x % 2; i++) ans += a[i];
    cout << ans << endl;
    return;
  }
  long long x = n / m;
  long long ans = (x / 10) * (45ll);
  for (int i = 1; i <= x % 10; i++) ans += a[i];
  cout << ans << endl;
  return;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please formulate a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, m;
  cin >> n >> m;
  if (m > n or m % 10 == 0) {
    cout << 0 << endl;
    return;
  }
  long long a[12];
  for (int i = 1; i <= 10; i++) {
    a[i] = (i * m) % 10;
  }
  if (m % 2 == 0) {
    long long x = n / m;
    long long ans = (x / 5) * (20ll);
    for (int i = 1; i <= x % 5; i++) ans += a[i];
    cout << ans << endl;
    return;
  }
  if (m % 10 == 5) {
    long long x = n / m;
    long long ans = (x / 2) * (5ll);
    for (int i = 1; i <= x % 2; i++) ans += a[i];
    cout << ans << endl;
    return;
  }
  long long x = n / m;
  long long ans = (x / 10) * (45ll);
  for (int i = 1; i <= x % 10; i++) ans += a[i];
  cout << ans << endl;
  return;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```