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 MinMexCut {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int numCases = Integer.parseInt(sc.next());
        
        for (int i = 0; i < numCases; i++) {
            System.out.println(minmex(sc.next()));
        }
    }
    private static int minmex(String str) {
        int firstZero = -1;
        for (int i = 0; i < str.length(); i++) {
            if (str.substring(i,i+1).equals("0")) {
                firstZero = i;
                break;
            }    
        }
        if (firstZero == -1) return 0;
        
        int mexSum = 1;
        
        boolean foundNextOne = false;
        for (int i = firstZero+1; i < str.length(); i++) {
            if (str.substring(i,i+1).equals("1")) foundNextOne = true;
            else {
                if (foundNextOne) {
                    mexSum++;
                    foundNextOne = false;
                }
            }
            if (mexSum == 2) break;
        }
        
        return mexSum;
    }
}