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 FuckMyLife() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int gcd(int a, int b) {
  if (!a || !b) return a + b;
  return gcd(b, a % b);
}
long long CalcArithmeticAn(long long a1, long long n, long long d) {
  return a1 + (n - 1) * d;
}
long long CalcArithmeticSn1(long long a1, long long an, long long n) {
  return (n * (a1 + an)) / 2;
}
long long CalcArithmeticSn2(long long a1, long long n, long long d) {
  return (n * (2 * a1 + (n - 1) * d)) / 2;
}
double CalcGeometricAn(long long a, long long r, long long n) {
  return (a * pow(r, n - 1));
}
double CalcGeometricSn(long long a, long long r, long long n) {
  return (a * (1 - pow(r, n))) / (1 - r);
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int a[100];
int main() {
  FuckMyLife();
  int tc;
  cin >> tc;
  while (tc--) {
    int a, b, n;
    cin >> a >> b >> n;
    if (a < b) swap(a, b);
    int cnt = 1;
    while (a <= n && b <= n) {
      if (cnt % 2)
        b += a;
      else
        a += b;
      cnt++;
    }
    cout << cnt - 1 << '\n';
  }
  return 0;
}
