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.
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 / self.d < other.n / other.d


class Interval:
    def __init__(self, start, end):
        self.start = start
        self.end = end

    def __str__(self):
        return str(self.start) + str(self.end)

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

    def __eq__(self, other):
        return self.start == other.start and self.end == other.end

    def __lt__(self, other):
        return self.start < other.start


def overlap(verl1, verl2):
    if verl1.end >= verl2.start:
        return True
    else:
        return False


def merge(verl1, verl2):
    retverl = Interval(min(verl1.start, verl2.start), max(verl1.end, verl2.end))
    return retverl


def changeBase(num, base):
    if base == 2:
        return "{0:b}".format(num)
    power = math.floor(math.log(num, base))
    remainder = num
    end = []
    iter = list(range(power + 1))
    iter.reverse()
    for i in iter:
        if remainder >= base ** i:
            q, r = divmod(remainder, base ** i)
            end.append(q)
            remainder = r
        else:
            end.append(0)
    retlist = []
    for i in end:
        if i > 9:
            retlist.append(uabet[i - 10])
        else:
            retlist.append(i)
    retlist = map(str, retlist)
    return ''.join(retlist)


def prime(num):
    if num == 2:
        return True
    if num % 2 == 0:
        return False
    for i in range(2, math.ceil(num ** 0.5) + 1):
        if num % i == 0:
            return False
    return True


def solve(s):
    z = s.count('0')
    o = s.count('1')
    if z==0:
        return 0
    if o == 0:
        return 1
    x = s.split('1')
    c= 0
    for  i in x:
        if i!='':
            c+=1
    if c<2:
        return c
    return 2

def main():
    t = int(input())
    for i in range(t):
        s = input()
        print(solve(s))


main()
