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 os,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)
try:sys.stdin,sys.stdout=open('in.txt','r'),open('out.txt','w')
except:pass
ii1=lambda:int(sys.stdin.readline().strip()) # for interger
is1=lambda:sys.stdin.readline().strip() # for str
iia=lambda:list(map(int,sys.stdin.readline().strip().split())) # for List[int]
isa=lambda:sys.stdin.readline().strip().split() # for List[str]
mod=int(1e9 + 7);
# from math import *
# from math import inf 
# from collections import *;
# from collections import defaultdict as dd
# from collections import deque as dq 
# from string import ascii_lowercase,ascii_uppercase
# from functools import lru_cache
# sys.setrecursionlimit(500000)
# from bisect import bisect_left as bl,bisect_right as br
# from heapq import heapify,heappop,heappush
# def google(p):print('Case #'+str(p)+': ',end='')
###################### Start Here ######################
from math import ceil
for _ in range(ii1()):
    s = is1()
    z = s.count('0')
    o = s.count('1')
    if z == 0 :
        print(0)
    elif o == 0:
        print(1)
    else:
        # 000....1111... ho to iska 1 ho sakta hai 
        i = 0 
        n = len(s)
        o1 = 0 
        while i < n and s[i]=='1':
            o1 += 1
            i += 1
        z1 = 0 
        while i < n  and s[i]=='0':
            z1 += 1
            i += 1
        o2 = 0
        while i < n and s[i]=='1':
            o2+=1
            i+=1
        if z1 + o1 + o2 == n:
            print(1)
        else:
            print(2)

        