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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long i, j, k, n, prev = 1, temp = 1, a = 1, b = 1, c, t, ans;
  vector<long long> vec, choto, boro;
  vec.push_back(1);
  vec.push_back(1);
  for (long long i = 0; i < 100; i++) {
    c = vec[i] + vec[i + 1];
    vec.push_back(c);
  }
  cin >> t;
  boro.push_back(0);
  choto.push_back(1);
  choto.push_back(0);
  for (long long i = 0; i < 100; i++) {
    c = vec[i];
    boro.push_back(c);
    choto.push_back(c);
  }
  while (t--) {
    cin >> a >> b >> n;
    if (b > a) swap(a, b);
    if (n < a)
      cout << "0"
           << "\n";
    else {
      for (long long i = 0; i < 100; i++) {
        temp = boro[i] * a + choto[i] * b;
        if (temp > n) {
          ans = i - 1;
          break;
        }
      }
      cout << ans << "\n";
    }
  }
}
