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>
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    if (a > c || b > c) {
      printf("0\n");
      continue;
    }
    int d = a, e = b;
    int ans1 = 0, ans2 = 0;
    for (int i = 0;;) {
      if (d > c || e > c) {
        ans1 = i;
        break;
      }
      d += e;
      i++;
      if (d > c || e > c) {
        ans1 = i;
        break;
      }
      e += d;
      i++;
    }
    d = a, e = b;
    for (int i = 0;;) {
      if (d > c || e > c) {
        ans2 = i;
        break;
      }
      e += d;
      i++;
      if (d > c || e > c) {
        ans2 = i;
        break;
      }
      d += e;
      i++;
    }
    int ans3 = (c - a) / b + 1;
    int ans4 = (c - b) / a + 1;
    int M = ans1;
    M = M < ans2 ? M : ans2;
    M = M < ans3 ? M : ans3;
    M = M < ans4 ? M : ans4;
    printf("%d\n", M);
  }
}
