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;
int main() {
  int t, a, b, x, y, z, n, mx, mn, tmp, a2, b2, a3, b3, a4, b4;
  scanf("%d", &t);
  for (int i = 0; i < t; i++) {
    int cnt = 0, ans = 0;
    string s = "";
    vector<int> vc;
    map<int, int> mp;
    map<char, int> mc;
    scanf("%d %d %d", &a, &b, &n);
    char ch, ch2;
    if (a > b) swap(a, b);
    a2 = a, b2 = b, a3 = a, b3 = b, a4 = a, b4 = b;
    ans = 0;
    while (1) {
      if (a > n || b > n || a2 > n || b2 > n || a3 > n || b3 > n || a4 > n ||
          b4 > n)
        break;
      ans++;
      a3 = a3 + b3;
      b4 = b4 + a4;
      if (ans % 2 == 0) {
        a = a + b;
        b2 = a2 + b2;
      } else {
        b = a + b;
        a2 = a2 + b2;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
