Codeforces Round #739 (Div. 3)--D

题干：

You are given an integer nn. In 11 move, you can do one of the following actions:

• erase any digit of the number (it's acceptable that the number before the operation has exactly one digit and after the operation, it is "empty");
• add one digit to the right.

The actions may be performed in any order any number of times.

Note that if, after deleting some digit from a number, it will contain leading zeroes, they will not be deleted. E.g. if you delete from the number 301301 the digit 33, the result is the number 0101 (not 11).

You need to perform the minimum number of actions to make the number any power of 22 (i.e. there's an integer kk (k≥0k≥0) such that the resulting number is equal to 2k2k). The resulting number must not have leading zeroes.

E.g. consider n=1052n=1052. The answer is equal to 22. First, let's add to the right one digit 44 (the result will be 1052410524). Then let's erase the digit 55, so the result will be 10241024 which is a power of 22.

E.g. consider n=8888n=8888. The answer is equal to 33. Let's erase any of the digits 88 three times. The result will be 88 which is a power of 22.

Input

The first line contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases. Then tt test cases follow.

Each test case consists of one line containing one integer nn (1≤n≤1091≤n≤109).

Output

For each test case, output in a separate line one integer mm — the minimum number of moves to transform the number into any power of 22.

Example

input

Copy

12
1052
8888
6
75
128
1
301
12048
1504
6656
1000000000
687194767

output

Copy

2
3
1
3
0
0
2
1
3
4
9
2

Note

The answer for the first test case was considered above.

The answer for the second test case was considered above.

In the third test case, it's enough to add to the right the digit 44 — the number 66 will turn into 6464.

In the fourth test case, let's add to the right the digit 88 and then erase 77 and 55 — the taken number will turn into 88.

The numbers of the fifth and the sixth test cases are already powers of two so there's no need to make any move.

In the seventh test case, you can delete first of all the digit 33 (the result is 0101) and then the digit 00 (the result is 11).

**思路：**

处理出2的n次幂t，将输入数字和t都在转成字符串匹配.（注意：n要尽可能的大）

代码：

int main(){
    int t;
    cin>>t;string s,temp;
    while(t--){
        cin>>s; int ans=20;
        for(int i=0;i<=60;i++){
            temp=to_string(1ll<<i);int res=0;
            int k=0;
            for(int j=0;j<s.size();j++){
                if(k<temp.size()&&s[j]==temp[k])k++;
                else res++;
            }
            res+=temp.size()-k;
            ans=min(ans,res);
        }
        cout<<ans<<endl;
    }
    return 0;
}