Leo has developed a new programming language C+=. In C+=, integer variables can only be changed with a "+=" operation that adds the right-hand side value to the left-hand side variable. For example, performing "a += b" when a = 2, b = 3 changes the value of a to 5 (the value of b does not change).

In a prototype program Leo has two integer variables a and b, initialized with some positive values. He can perform any number of operations "a += b" or "b += a". Leo wants to test handling large integers, so he wants to make the value of either a or b strictly greater than a given value n. What is the smallest number of operations he has to perform?

Input

The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.

Each of the following T lines describes a single test case, and contains three integers a, b, n (1 ≤ a, b ≤ n ≤ 10^9) — initial values of a and b, and the value one of the variables has to exceed, respectively.

Output

For each test case print a single integer — the smallest number of operations needed. Separate answers with line breaks.

Example

Input


2
1 2 3
5 4 100


Output


2
7

Note

In the first case we cannot make a variable exceed 3 in one operation. One way of achieving this in two operations is to perform "b += a" twice.
import java.util.Scanner;
import java.util.Arrays;


public class CT {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int [] ar=new int[100000];

        int t = s.nextInt();
        //   int tt=t;
        for (int pp=0;pp<t;pp++) {

            int a = s.nextInt();
            int b = s.nextInt();
            int n = s.nextInt();
            int aa, bb;

            aa = Math.max(a, b);
            bb = Math.min(a, b);
            int cc = 0;

            while (bb <= n) {
                int x=aa,y=bb;
                aa = Math.max(x, y);
                bb = Math.min(x, y);


                if ((bb > n)||(aa>n)) break;
                else {
                    bb += aa;
                    cc++;

                }
            }


                System.out.println(cc);



        }

        //for (int pp=0;pp<t;pp++){
          //  int kk=0;

           // System.out.println(ar[pp]);
      //  }


}}