Yet Another Meme Problem

You are given two integers $A$ and $B$, calculate the number of pairs $(a,b)$ such that $1\le a\le A,1\le b\le B$, and the equation $a\cdot b+a+b=conc(a,b)$ 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 $t(1\le t\le 100)$ — the number of test cases.

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

Output

Print one integer — the number of pairs $(a,b)$ such that $1\le a\le A,1\le b\le B$, and the equation $a\cdot b+a+b=conc(a,b)$ 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: $a=1,b=9(1+9+1\cdot 9=19)$.

#include<bits/stdc++.h>
 
using namespace std;
typedef long long ll;
ll A,B;
ll b[15];
int main() {
    
    
    ll c = 1;
    for (int i = 1; i <= 10; i++) {
    
    
        c = c * 10;
        b[i] = c - 1;
    }
    int _;
    scanf("%d",&_);
    while (_--) {
    
    
        scanf("%lld%lld", &A, &B);
        int i = 1;
        while (b[i] <= B) {
    
    
            i++;
        }
        printf("%lld\n", A * (i - 1));
    }
    return 0;
}