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 long long INF = 1e12;
const long long MOD = 998244353;
const long long mod = 1e9 + 7;
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (T i : v) os << i << ' ';
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const set<T>& v) {
  for (T i : v) os << i << ' ';
  return os;
}
template <class T, class S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << v.first << ' ' << v.second;
  return os;
}
template <class T, class S>
ostream& operator<<(ostream& os, const unordered_map<T, S>& v) {
  for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' ';
  return os;
}
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target( \
    "avx2,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
clock_t time_p = clock();
void starboy1299() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
inline long long mul(long long a, long long b, long long m = mod) {
  return (a * b) % m;
}
inline long long add(long long a, long long b, long long m = mod) {
  return (a + b) % m;
}
inline long long sub(long long a, long long b, long long m = mod) {
  return (a - b + m) % m;
}
const long long N = 2e5 + 5;
void solve() {
  long long a, b, n;
  cin >> a >> b >> n;
  if (a > n or b > n) {
    cout << 0 << '\n';
    return;
  }
  long long cnt = 0;
  while (1) {
    if (a < b) {
      a += b;
    } else
      b += a;
    cnt++;
    if (a > n or b > n) {
      break;
    }
  }
  cout << cnt << '\n';
}
int32_t main() {
  fast();
  long long t;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  starboy1299();
}
