human
PYTHON3 solution for "You are given an integer n. In 1 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 301 the digit 3, the result is the number 01 (not 1).

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

E.g. consider n=1052. The answer is equal to 2. First, let's add to the right one digit 4 (the result will be 10524). Then let's erase the digit 5, so the result will be 1024 which is a power of 2.

E.g. consider n=8888. The answer is equal to 3. Let's erase any of the digits 8 three times. The result will be 8 which is a power of 2.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

Each test case consists of one line containing one integer n (1 ≤ n ≤ 10^9).

Output

For each test case, output in a separate line one integer m — the minimum number of moves to transform the number into any power of 2.

Example

Input

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

Output

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 4 — the number 6 will turn into 64.

In the fourth test case, let's add to the right the digit 8 and then erase 7 and 5 — the taken number will turn into 8.

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...":



Can you provide a the Python code you requested?
assistant
#Here is the Python code you requested.
 # ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0
    
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
    
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# ------------------- fast io --------------------
from math import ceil

def prod(a, mod=10 ** 9 + 7):
    ans = 1
    for each in a:
        ans = (ans * each) % mod
    return ans

def gcd(x, y):
    while y:
        x, y = y, x % y
    return x

def lcm(a, b):...