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 err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ","
       << " ";
  err(++it, args...);
  cout << "\n";
}
bool checkbit(long long mask, long long bit) { return mask & (1LL << bit); }
int setbit(int mask, int bit) { return mask | (1 << bit); }
int clearbit(int mask, int bit) { return mask & ~(1 << bit); }
int togglebit(int mask, int bit) { return mask ^ (1 << bit); }
int bitno(int mask) { return (int)__builtin_popcount(mask); }
int fx[] = {0, 0, +1, -1};
int fy[] = {+1, -1, 0, 0};
int x_move[] = {2, 1, -1, -2, -2, -1, 1, 2};
int y_move[] = {1, 2, 2, 1, -1, -2, -2, -1};
const int maX = 1e3 + 5;
const int maXC = 55;
map<string, long long> check;
void solve() {
  long long n, i, j, cc = 0, a, b;
  cin >> a >> b >> n;
  if (a < b) swap(a, b);
  while (a <= n) {
    cc++;
    long long x = a;
    a += b;
    b = x;
  }
  cout << cc << endl;
}
int main() {
  int test;
  test = 1;
  scanf("%d", &test);
  while (test--) solve();
  return 0;
}
