#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int N = 1e6 + 10;
const long long mod = 1e9 + 7;
int dig[12];
int main() {
  int t;
  long long n, m;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &n, &m);
    long long sum = 0;
    for (int i = 1; i <= 10; i++) dig[i - 1] = (i * m) % 10, sum += dig[i - 1];
    long long ans = 0;
    n /= m;
    long long len = 0;
    for (int i = 0; i < n % 10; i++) len += dig[i];
    ans = len + n / 10 * sum;
    printf("%lld\n", ans);
  }
}

### Prompt

In CPP, your task is to solve 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;
const int maxn = 1e5 + 10;
const int N = 1e6 + 10;
const long long mod = 1e9 + 7;
int dig[12];
int main() {
  int t;
  long long n, m;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &n, &m);
    long long sum = 0;
    for (int i = 1; i <= 10; i++) dig[i - 1] = (i * m) % 10, sum += dig[i - 1];
    long long ans = 0;
    n /= m;
    long long len = 0;
    for (int i = 0; i < n % 10; i++) len += dig[i];
    ans = len + n / 10 * sum;
    printf("%lld\n", ans);
  }
}

```