#include <bits/stdc++.h>
using namespace std;
 
void solve(){
	long long a ,b ;
	cin>>a>>b;
 
	int op = 0;
	int minops = INT_MAX;
 
	// for the worst case scenario 10^9  and 1 we will need 30 max oprations log2(10^9) = 30 
	// dividing takes O(logn) time;
 
	// so what we will do is that we will iterate b for 30 times and will try to find combination of
	//type 1 and type 2 oprations so that it will take min 
 
 
	for(int i =0; i < 30 ; i++){
		long long tempa = a;
		long long tempb = b+i;
 
		if(tempb == 1) continue;
 
		int op = i;
		while(tempa>0){
			tempa=tempa/tempb;
			op++;
		}
		minops = min(minops , op);
	}
 
	cout<<minops<<endl;
 
	// while(a>0){
	// 	b++;
	// 	a=a/b;
	// 	op++;
 
	// 	minops = min(minops , op);
	// }
 
 
}
 
 
 
int main() {
	int t;
	cin>>t;
	while(t--){
		solve();
	}
	return 0;
}

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6
9 2
1337 1
1 1
50000000 4
991026972 997
1234 5678