#include <bits/stdc++.h>
using namespace std;
void flash() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long pw(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long ans = pw(x, y / 2);
  ans = ans * ans;
  if (y % 2 == 1) ans = ans * x;
  return ans;
}
long long pwm(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long ans = pwm(x, y / 2) % 998244353;
  ans = (ans * ans) % 998244353;
  if (y % 2 == 1) ans = (ans * x) % 998244353;
  return ans;
}
long long check(long long n) {
  if (n < 0)
    return 0;
  else
    return 1;
}
long long add(long long a, long long b) { return a + b; }
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
long long minv(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m, t = m;
    m = a % m, a = t, t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
bool cmp(long long i1, long long i2) { return (i1 > i2); }
void solve() {
  long long a, b, n;
  cin >> a >> b >> n;
  long long ans = 0;
  while (a <= n and b <= n) {
    if (a > b) {
      b += a;
    } else {
      a += b;
    }
    ans++;
  }
  cout << ans << "\n";
}
int32_t main() {
  flash();
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void flash() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long pw(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long ans = pw(x, y / 2);
  ans = ans * ans;
  if (y % 2 == 1) ans = ans * x;
  return ans;
}
long long pwm(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long ans = pwm(x, y / 2) % 998244353;
  ans = (ans * ans) % 998244353;
  if (y % 2 == 1) ans = (ans * x) % 998244353;
  return ans;
}
long long check(long long n) {
  if (n < 0)
    return 0;
  else
    return 1;
}
long long add(long long a, long long b) { return a + b; }
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
long long minv(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m, t = m;
    m = a % m, a = t, t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
bool cmp(long long i1, long long i2) { return (i1 > i2); }
void solve() {
  long long a, b, n;
  cin >> a >> b >> n;
  long long ans = 0;
  while (a <= n and b <= n) {
    if (a > b) {
      b += a;
    } else {
      a += b;
    }
    ans++;
  }
  cout << ans << "\n";
}
int32_t main() {
  flash();
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```