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;
using ll = long long;
using ld = long double;
using pr = pair<int, int>;
using prll = pair<long, long>;
using V = vector<int>;
using Vll = vector<long long>;
template <typename T, typename TT>
pr operator+(pair<T, TT> a, pair<T, TT> b) {
  return {a.first + b.first, a.second + b.second};
}
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (auto itr : t) s << itr << " ";
  return s;
}
ll a, b, n;
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> a >> b >> n;
    if (a > b) swap(a, b);
    int ans = 0;
    while (a <= n && b <= n) {
      a += b;
      ans++;
      if (a > n) break;
      b += a;
      ans++;
    }
    cout << ans << "\n";
  }
  return 0;
}
