Try guessing the statement from this picture <http://tiny.cc/ogyoiz>.

You are given two integers A and B, calculate the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true; conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). a and b should not contain leading zeroes.

Input

The first line contains t (1 ≤ t ≤ 100) — the number of test cases.

Each test case contains two integers A and B (1 ≤ A, B ≤ 10^9).

Output

Print one integer — the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true.

Example

Input


3
1 11
4 2
191 31415926


Output


1
0
1337

Note

There is only one suitable pair in the first test case: a = 1, b = 9 (1 + 9 + 1 ⋅ 9 = 19).
#include <bits/stdc++.h>
using namespace std;
const int next_x[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const int next_y[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const int inf = 2147483647;
const long long linf = 1e18 + 5;
const double PI = acos(-1.0);
const long double eps = 1e-12;
const int MAXN = 20 + 5;
const int N = 4e6 + 10;
const int fix = 1e5;
long long n, d;
int con(int i, int j) {
  int x = j;
  while (x) x /= 10, i *= 10;
  return i + j;
}
int work() {
  int CASE;
  scanf("%d", &CASE);
  while (CASE--) {
    long long n, m;
    scanf("%lld%lld", &n, &m);
    long long ans = 0, t = 0;
    long long cnt = 0;
    while (t * 10 + 9 <= m) t = t * 10 + 9, cnt++;
    printf("%lld\n", n * cnt);
  }
  return 0;
}
int main() {
  work();
  return 0;
}
