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 MOD = 1e9 + 7;
const int MAXN = 1e5 + 50;
long long qpow(long long a, long long b) {
  long long ret = 1;
  b %= MOD - 1;
  a %= MOD;
  while (b) {
    if (b & 1) ret = a * ret % MOD;
    b >>= 1;
    a = a * a % MOD;
  }
  return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void work() {
  long long a, b;
  scanf("%lld%lld", &a, &b);
  long long ans = 0;
  int ned = 9;
  while (b >= ned) {
    ans += a;
    ned = ned * 10 + 9;
  }
  printf("%lld\n", ans);
}
int main() {
  int T;
  scanf("%d", &T);
  for (int cas = 1; cas <= T; cas++) work();
}
