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;
const int N = 1e5 + 5;
const long long M = (long long)1e9 + 7;
const double pi = 2 * acos(0.0);
const double esp = 1e-9;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  cin >> test;
  for (int cs = 1; cs <= test; ++cs) {
    long long a, b, n, cnt = 0, a1, b1;
    cin >> a1 >> b1 >> n;
    if (n < a1 || n < b1) {
      cout << 0 << endl;
    } else {
      long long ans = INT_MAX;
      a = a1, b = b1;
      while (1) {
        a += b;
        cnt++;
        if (a > n) break;
        b += a;
        cnt++;
        if (b > n) break;
      }
      ans = min(ans, cnt);
      cnt = 0;
      a = a1, b = b1;
      while (1) {
        b += a;
        cnt++;
        if (b > n) break;
        a += b;
        cnt++;
        if (a > n) break;
      }
      ans = min(ans, cnt);
      cout << ans << endl;
    }
  }
  return 0;
}
