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.io.*;
import java.util.*;

public class c {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine().trim());
        int t = Integer.parseInt(st.nextToken());

        for (int e = 1; e <= t; e++) {
            st = new StringTokenizer(br.readLine());
            long a = Long.parseLong(st.nextToken());
            long b = Long.parseLong(st.nextToken());
            long n = Long.parseLong(st.nextToken());
            if (Math.max(a, b) > n)
                System.out.println("0");
            else {
                if(a>b)
                {
                    long temp=a;
                    a=b;
                    b=temp;
                }
                long sum = a + b;
                int count = 1;
                while (sum <= n) {
                    a = b;
                    b = sum;
                    sum = a + b;
                    count++;
                }
                System.out.println(count);
            }
        }
    }
}