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 os

import sys

debug = True

if debug and os.path.exists("input.in"):
    input = open("input.in", "r").readline
else:
    debug = False
    input = sys.stdin.readline


def inp():
    return (int(input()))


def inlt():
    return (list(map(int, input().split())))


def insr():
    s = input()
    return s[:len(s) - 1]  # Remove line char from end


def invr():
    return (map(int, input().split()))


test_count = int(input())

for t in range(test_count):
    if debug:
        print("Test Case #", t + 1)
    # Start code here
    a, b, n = invr()
    if a > b:
        a, b = b, a
    if a + b > n:
        print(1)
    else:
        max_f = n // (a + b)
        f = list()
        x1 = 1
        x2 = 1
        f.append(x1)
        f.append(x2)
        while x2 <= max_f:
            f.append(x1 + x2)
            x1, x2 = x2, x1 + x2
        # print(f)
        ans = len(f)
        for i in range(len(f) - 1, -1, -1):
            # print(i, f[i] * b + f[i - 1] * a)
            if (f[i] * b + f[i - 1] * a) <= n:
                break
            ans = i
        print(ans)
