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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2")
using namespace std;
using lld = long long int;
using pll = pair<lld, lld>;
using lf = float;
using pff = pair<lf, lf>;
using pii = pair<int, int>;
int n, m;
lld cal(lld p, lld q, lld r) {
  int type = 0;
  int cnt = 0;
  while (p <= r && q <= r) {
    if (type == 0)
      p += q;
    else
      q += p;
    type ^= 1;
    cnt++;
  }
  return cnt;
}
int main() {
  int i, j, k;
  int t = 1, tv = 0;
  scanf("%d", &t);
  while (t--) {
    lld p, q, r;
    scanf("%lld %lld %lld", &p, &q, &r);
    lld s = cal(p, q, r);
    lld t = cal(q, p, r);
    printf("%lld\n", min(s, t));
  }
}
