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.
# cook your dish here
# cook your dish here
#import math as m
import sys
try:
    
    for t in range(int(sys.stdin.readline())):
         #n = int(sys.stdin.readline())
         #a = list(map(int,sys.stdin.readline().split()))
         #b = list(map(int,sys.stdin.readline().split()))
         s = sys.stdin.readline()
         #n,s = map(int,sys.stdin.readline().split())
         c_1 = s.count('1')
         c_0 = s.count('0')
         if(c_0==0):
              print(0)
         elif(c_1==0):
              print(1)
         elif(c_1>=0 and c_0==1):
              print(1)
         else:
              cu_1 = 0 
              cu_0 = 0 
              for i in range(1,len(s)-1):
                   if(s[i]=='1'):
                        if(s[i]==s[i-1]):
                             pass 
                        else:
                             cu_0 = cu_0 +1 
                   elif(s[i]=='0'):
                        if(s[i]==s[i-1]):
                             pass
                        else:
                             cu_1 = cu_1 + 1 
              if(s[len(s)-2]=='0'):
                   cu_0 = cu_0+1 
              else:
                   cu_1 = cu_1 +1 
              #print(s[len(s)-2], cu_1 ,cu_0)
              if(cu_0==0):
                   print(0)
              elif(cu_1==0):
                   print(1)
              elif(cu_1>=1 and cu_0==1):
                   print(1)
                   
              else:     
                   print(2)
         
        
         
         #sys.stdout.write(str()+'\n')
         #print(res)
                   
              
except EOFError as e:
    pass
