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.

// Problem: A. Dense Array
// Contest: Codeforces - Codeforces Round #702 (Div. 3)
// URL: https://codeforces.com/contest/1490/problem/A
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define ll          long long
#define pb          push_back
#define eb 			emplace_back
#define mp 			make_pair
#define pii         pair<ll,ll>
#define vi          vector<int>
#define vii         vector<pii>
#define mi          map<ll,ll>
#define mii         map<pii,ll>
#define all(a)      (a).begin(),(a).end()
#define fi          first
#define si          second
#define sz(x)       (ll)x.size()
#define endl        "\n"
#define checkbit(n, b) ((n >> b) & 1)
#define mod        1000000007
#define inf        1000000000000000000
#define puts(n)		cout<<n;
#define um           unordered_map<ll,ll>
#define rep(i,a,b)  for(int i=a;i<b;i++)
#define setbits(x) __builtin_popcount(x)
#define mset(m,v) memset(m,v,sizeof(m))
#define int ll 
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
using namespace std;
int power(int base, int exp); 
 
//=======================
 
 // USED UNORDERED_MAP????????
 // INT_MAX and LONG_MAX range???
 //gcd(x,y)=gcd(x-y,y)
 //gcd(x,y,z...)=gcd(x,y-x,z-x,...)
void solve() {
 int n;
 cin>>n;
 vi arr(n,0);
 rep(i,0,n)
 {
 	cin>>arr[i];
 }
 int cnt=0;
 rep(i,0,n-1)
 {
 	int x=max(arr[i],arr[i+1]);
 	int y=min(arr[i],arr[i+1]);
 	while(y*2<x)
 	{
 		x=(x+1)/2;
 		cnt++;
 	}
 }
 cout<<cnt<<endl;
}
 
int32_t main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    srand(chrono::high_resolution_clock::now().time_since_epoch().count());
 
   int t = 1;
    cin >> t;
    while(t--) {
      solve();
    }
 
    return 0;
}
 
int power(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((ll)result * base) % mod;
    base = ((ll)base * base) % mod;
    exp >>= 1;
  }
  return result;
}