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;
void solve() {
  long long n, a, b, i, ans;
  cin >> a >> b >> n;
  ans = 0;
  if (a < b) swap(a, b);
  for (i = 0;; i++) {
    if (i % 2)
      a += b;
    else
      b += a;
    ans++;
    if (a > n or b > n) break;
  }
  cout << ans << '\n';
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  cerr << " Execution : " << (1.0 * clock()) / CLOCKS_PER_SEC << "s \n";
  return 0;
}
