/*******************************************************
Problem - Factorials and exponents
Solution by - Sean Falconer
Algorithm: 
  The solution involves a little math.  
  Given n! >= m^d, we want to find a valid d.
  
  If we take the log of both sides we get:
  log(n!) >= d*log(m)
  
  However, n! = n*(n-1)*(n-2)*...*1, so
  log(n!) = log(n) + log(n-1) + ... + log(1).
  
  Thus, we can rewrite the equation as:
  log(n) + log(n-1) + ... + log(1) >= d*log(m).
  
  Now solve for d and we get:
  (log(n) + log(n-1) + ... + log(1)) / log(m) >= d.

  Now, we just need to compute the first part in a loop
  and divide by m.  		   
*******************************************************/

#include <iostream>
#include <math.h>
using namespace std;

const double EPS = 1e-5;

int main() {
  int T;
  cin >> T;
  for(int i = 0; i < T; i++) {
    int n; double m;
    cin >> n >> m;

    double logfac = 0;
    for(int j = 1; j <= n; j++) logfac += log((double)j);

    logfac /= log(m);

    if(fabs(m - 1.0) < EPS || m < 1.0) cout << "infinity" << endl;
    else cout << (int)logfac << endl;
  }
  return 0;
}