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.
#include <bits/stdc++.h>
using namespace std;
long long numbit(long long x) { return (1LL << x) - 1LL; }
long long getbit(long long x, long long i) { return (x >> i) & 1LL; }
long long onbit(long long x, long long i) { return x | (1LL << i); }
long long offbit(long long x, long long i) { return x & (~(1LL << i)); }
const int oo = 1000111000;
const int Nmax = 1e5 + 10;
const long long mod = 1e9 + 7;
const long double PI = 3.14159265358979323846;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int e;
  cin >> e;
  for (long long z = (1); z <= (e); z++) {
    long long x, y, n;
    cin >> x >> y >> n;
    if (x > y) swap(x, y);
    long long r = 0;
    while (y <= n) {
      long long z = (x + y);
      x = y;
      y = z;
      r++;
    }
    cout << r << endl;
  }
}
