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 java.util.*;
public class Solution {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
              Scanner sc=new Scanner(System.in);
              int t =sc.nextInt();
              while(t-->0) {
            	String str =sc.next();
            	int n=str.length();
            	int count1=0;
            	int count2=0;
            	int count3=0;
            	for(int i=0;i<n;i++) {
            
            		if(str.charAt(i)=='0') {
            			count1++;
            		}else if(str.charAt(i)=='1') {
            			count2++;
            		}else {
            			count3++;
            		}
            			
            			
            			
            	}
            	if(count1 ==0) {
            		System.out.println("0");
            	}else if(count2==0) {
            		System.out.println("1");
            	}else if (count3==0) {
            		if(count1<2) {
            			System.out.println(count1);
            		}else {
            			System.out.println("2");
            		}
            	
            	}else {
            		System.out.println(count1);
            	}
            	
              }
	}

}