Codeforces Round #647 (Div. 2)

Johnny has recently found an ancient, broken computer. The machine has only one register, which allows one to put in there one variable. Then in one operation, you can shift its bits left or right by at most three positions. The right shift is forbidden if it cuts off some ones. So, in fact, in one operation, you can multiply or divide your number by 22 , 44 or 88 , and division is only allowed if the number is divisible by the chosen divisor.

Formally, if the register contains a positive integer xx , in one operation it can be replaced by one of the following:

• x⋅2x⋅2
• x⋅4x⋅4
• x⋅8x⋅8
• x/2x/2 , if xx is divisible by 22
• x/4x/4 , if xx is divisible by 44
• x/8x/8 , if xx is divisible by 88

For example, if x=6x=6 , in one operation it can be replaced by 1212 , 2424 , 4848 or 33 . Value 66 isn't divisible by 44 or 88 , so there're only four variants of replacement.

Now Johnny wonders how many operations he needs to perform if he puts aa in the register and wants to get bb at the end.

Input

The input consists of multiple test cases. The first line contains an integer tt (1≤t≤10001≤t≤1000 ) — the number of test cases. The following tt lines contain a description of test cases.

The first and only line in each test case contains integers aa and bb (1≤a,b≤10181≤a,b≤1018 ) — the initial and target value of the variable, respectively.

Output

Output tt lines, each line should contain one integer denoting the minimum number of operations Johnny needs to perform. If Johnny cannot get bb at the end, then write −1−1 .

Example

Input

Copy

10
10 5
11 44
17 21
1 1
96 3
2 128
1001 1100611139403776
1000000000000000000 1000000000000000000
7 1
10 8

Output

Copy

1
1
-1
0
2
2
14
0
-1
-1
首先乘除互为逆运算，因此可以使a为较小的数，b为较大的数，全部转化为乘。
然后看a能否通过数次*2得到b，不能的话输出-1，否则贪心地先选择*8..来凑次数。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        long long a,b;
        int ans=0;
        cin>>a>>b;
        if(a>b) swap(a,b);
        bool flag=0;
        int cnt=0;
        while(a<b)
        {
            cnt++;
            a*=2;
        }
        if(a>b)
        {
            cout<<-1<<endl;
            continue;
        }
        else
        {
            ans+=cnt/3;
            cnt%=3;
            ans+=cnt/2;
            cnt%=2;
            ans+=cnt;
        }
        cout<<ans<<endl;
    }
    return 0;
}