human
Can you solve the below in PYTHON3?
A binary string is a string that consists of characters 0 and 1.

Let \operatorname{MEX} of a binary string be the smallest digit among 0, 1, or 2 that does not occur in the string. For example, \operatorname{MEX} of 001011 is 2, because 0 and 1 occur in the string at least once, \operatorname{MEX} of 1111 is 0, because 0 and 2 do not occur in the string and 0 < 2.

A binary string s is given. You should cut it into any number of substrings such that each character is in exactly one substring. It is possible to cut the string into a single substring — the whole string.

A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

What is the minimal sum of \operatorname{MEX} of all substrings pieces can be?

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.

Each test case contains a single binary string s (1 ≤ |s| ≤ 10^5).

It's guaranteed that the sum of lengths of s over all test cases does not exceed 10^5.

Output

For each test case print a single integer — the minimal sum of \operatorname{MEX} of all substrings that it is possible to get by cutting s optimally.

Example

Input

6
01
1111
01100
101
0000
01010

Output

1
0
2
1
1
2

Note

In the first test case the minimal sum is \operatorname{MEX}(0) + \operatorname{MEX}(1) = 1 + 0 = 1.

In the second test case the minimal sum is \operatorname{MEX}(1111) = 0.

In the third test case the minimal sum is \operatorname{MEX}(01100) = 2.

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Can you provide a Python program to print a single integer — the minimal sum of \operatorname{MEX} of all substrings that it is possible to get by cutting s optimally?
assistant
#Here is Python program to print a single integer — the minimal sum of \operatorname{MEX} of all substrings that it is possible to get by cutting s optimally.

import copy
import math

fin = None
fout = None
uabet = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V',
         'W', 'X', 'Y', 'Z']
abet = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v',
        'w', 'x', 'y', 'z']
INF = 10 ** 18

def start(filename):
    filename = filename.replace('.py', '')
    global fin
    global fout
    infile = filename + ".in"
    outfile = filename + ".out"
    fin = open(infile)
    fout = open(outfile, "w")

def permutation(lst):
    if len(lst) == 0:
        return []
    if len(lst) == 1:
        return [lst]
    l = []
    for i in range(len(lst)):
        m = lst[i]
        remLst = lst[:i] + lst[i + 1:]
        for p in permutation(remLst):
            l.append([m] + p)
    return l

def subsets(x):
    retset = []
    if len(x) == 0:
        retset = [[]]
    else:
        first = x[-1]
        cop = x.copy()
        cop.remove(first)
        y = subsets(cop)
        retset.extend(copy.deepcopy(y))
        m = copy.deepcopy(y)
        for i in range(len(y)):
            k = m[0]
            m.remove(k)
            k.append(first)
            m.append(k)
        retset.extend(m.copy())
    return retset

class Point:
    def __init__(self, x, y):
        self.x = x
        self.y = y

    def __str__(self):
        return str(self.x) + str(self.y)

    def __hash__(self):
        return hash(str(self))

    def __eq__(self, other):
        return self.x == other.x and self.y == other.y

class frac:
    def __init__(self, n, d):
        self.n = n
        self.d = d

    def __str__(self):
        return str(self.n) + '/' + str(self.d)

    def __hash__(self):
        return hash(str(self))

    def simp(self):
        div = math.gcd(self.n, self.d)
        self.n = self.n / div
        self.d = self.d / div

    def __eq__(self, other):
        return self.n / self.d == other.n / other.d

    def __lt__(self, other):
        return self.n /...