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 adskiy_razgon() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int mxN = 1e+9 + 7;
const long long INF = 1e+18 + 7;
long long nod(long long a, long long b) {
  if (b > a) {
    swap(a, b);
  }
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long nok(long long a, long long b) { return a * b / nod(a, b); }
void sp(long long a, double b) { cout << fixed << setprecision(a) << b; }
long long binpow(long long a, long long n) {
  if (n == 0) {
    return 1;
  }
  if (n % 2 == 1) {
    return binpow(a, n - 1) * a;
  } else {
    long long b = binpow(a, n / 2);
    return b * b;
  }
}
void solve() {
  long long a, b, c;
  cin >> a >> b >> c;
  long long ans = 0;
  while (a <= c && c >= b) {
    if (a > b) {
      b += a;
    } else {
      a += b;
    }
    ans++;
  }
  cout << ans << "\n";
}
int main() {
  adskiy_razgon();
  long long t = 1;
  cin >> t;
  for (int i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}
