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 int maxn = 2000000000;
vector<long long int> ve;
void fib() {
  ve.push_back(1);
  ve.push_back(1);
  for (long long int i = 2; ve[i - 1] < maxn; i++) {
    ve.push_back(ve[i - 1] + ve[i - 2]);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  fib();
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b, n;
    cin >> a >> b >> n;
    if (a > n || b > n) {
      cout << "0" << endl;
      continue;
    }
    long long int ca = a, cb = b;
    long long int i = 0, val = 0;
    while (val <= n) {
      val = ve[i] * a + ve[i + 1] * b;
      i++;
    }
    long long int j = 0, val1 = 0;
    while (val1 <= n) {
      val1 = ve[j] * cb + ve[j + 1] * ca;
      j++;
    }
    cout << min(i, j) << endl;
  }
  return 0;
}
