#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#define pb push_back
#define endl '\n';
#define salam  return 0
#define sz(s) int(s.size())
#define Ceil(x,y)  ((x+y-1)/y)
#define all(v) ((v).begin()),((v).end())
#define allr(v) ((v).rbegin()), ((v).rend()) 
#define mem(x,y) memset(x,y,sizeof(x))
#define fix(n)  fixed<<setprecision(n)
#define palindrom(s)  (s==string(s.rbegin(),s.rend()))
#define I64 int 
typedef long long ll;
typedef unsigned long long ull;
typedef double dl;
typedef long double ld;
const ll mod = ll(1e9 + 7);
const double PI = acos(-1.0);
const int N = (int)1e6 * 2, OO = 0x3f3f3f3f;
using namespace std;
void Selim()
{
	std::ios_base::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
}
int dx[8] = { 0, -1, 0, 1, -1, 1, -1, 1 };
int dy[8] = { 1, 0, -1, 0, -1, 1, 1, -1 };
ll gcd(ll a, ll b) { return !b ? a : gcd(b, a % b); }
ll LCM(ll a, ll b) { return a / gcd(a, b)*b; }
//ll Mod(ll x, ll m){ return ((x % m) + m) % m; }
ll fact(long long  n) { return !n ? 1 : n * fact(n - 1); }
//transform(s.begin(), s.end(), s.begin(), ::tolower);
//palindrom if(s==string(s.rbegin(),s.rend()))
//memset(arr,0,sizeof arr);   
int main()
{
	Selim();
	int t, n;
	cin >> t;
	while (t--){
		cin >> n;
		vector<int>v(n);
		for (int i = 0; i < n; i++){
			cin >> v[i];
		}
		int cnt = 0, cur = v[0];
		for (int i = 1; i < n; i++){
			if (max(cur, v[i])>2 * min(cur, v[i])){
				if (v[i]>cur){
					cur *= 2;
				}
				else{
					cur = Ceil(cur, 2);
				}
				cnt++; i--;
			}
			else{
				cur = v[i];
			}
		}
		cout << cnt << endl;
	}
	salam;
}
### Prompt

Develop a solution in Cpp to the problem described below:

Polycarp calls an array dense if the greater of any two adjacent elements is not more than twice bigger than the smaller. More formally, for any i (1 ≤ i ≤ n-1), this condition must be satisfied: $$$(max(a[i], a[i+1]))/(min(a[i], a[i+1])) ≤ 2$$$

For example, the arrays [1, 2, 3, 4, 3], [1, 1, 1] and [5, 10] are dense. And the arrays [5, 11], [1, 4, 2], [6, 6, 1] are not dense.

You are given an array a of n integers. What is the minimum number of numbers you need to add to an array to make it dense? You can insert numbers anywhere in the array. If the array is already dense, no numbers need to be added.

For example, if a=[4,2,10,1], then the answer is 5, and the array itself after inserting elements into it may look like this: a=[4,2,\underline{3},\underline{5},10,\underline{6},\underline{4},\underline{2},1] (there are other ways to build such a).

Input

The first line contains one integer t (1 ≤ t ≤ 1000). Then t test cases follow.

The first line of each test case contains one integer n (2 ≤ n ≤ 50) — the length of the array a.

The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50).

Output

For each test case, output one integer — the minimum number of numbers that must be added to the array to make it dense.

Example

Input


6
4
4 2 10 1
2
1 3
2
6 1
3
1 4 2
5
1 2 3 4 3
12
4 31 25 50 30 20 34 46 42 16 15 16


Output


5
1
2
1
0
3

Note

The first test case is explained in the statements.

In the second test case, you can insert one element, a=[1,\underline{2},3].

In the third test case, you can insert two elements, a=[6,\underline{4},\underline{2},1].

In the fourth test case, you can insert one element, a=[1,\underline{2},4,2].

In the fifth test case, the array a is already dense.

### Response

```cpp
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#define pb push_back
#define endl '\n';
#define salam  return 0
#define sz(s) int(s.size())
#define Ceil(x,y)  ((x+y-1)/y)
#define all(v) ((v).begin()),((v).end())
#define allr(v) ((v).rbegin()), ((v).rend()) 
#define mem(x,y) memset(x,y,sizeof(x))
#define fix(n)  fixed<<setprecision(n)
#define palindrom(s)  (s==string(s.rbegin(),s.rend()))
#define I64 int 
typedef long long ll;
typedef unsigned long long ull;
typedef double dl;
typedef long double ld;
const ll mod = ll(1e9 + 7);
const double PI = acos(-1.0);
const int N = (int)1e6 * 2, OO = 0x3f3f3f3f;
using namespace std;
void Selim()
{
	std::ios_base::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
}
int dx[8] = { 0, -1, 0, 1, -1, 1, -1, 1 };
int dy[8] = { 1, 0, -1, 0, -1, 1, 1, -1 };
ll gcd(ll a, ll b) { return !b ? a : gcd(b, a % b); }
ll LCM(ll a, ll b) { return a / gcd(a, b)*b; }
//ll Mod(ll x, ll m){ return ((x % m) + m) % m; }
ll fact(long long  n) { return !n ? 1 : n * fact(n - 1); }
//transform(s.begin(), s.end(), s.begin(), ::tolower);
//palindrom if(s==string(s.rbegin(),s.rend()))
//memset(arr,0,sizeof arr);   
int main()
{
	Selim();
	int t, n;
	cin >> t;
	while (t--){
		cin >> n;
		vector<int>v(n);
		for (int i = 0; i < n; i++){
			cin >> v[i];
		}
		int cnt = 0, cur = v[0];
		for (int i = 1; i < n; i++){
			if (max(cur, v[i])>2 * min(cur, v[i])){
				if (v[i]>cur){
					cur *= 2;
				}
				else{
					cur = Ceil(cur, 2);
				}
				cnt++; i--;
			}
			else{
				cur = v[i];
			}
		}
		cout << cnt << endl;
	}
	salam;
}
```