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.
/*

Sample Input 1

6
01
1111
01100
101
0000
01010

Sample Output 1

1
0
2
1
1
2

*/
import java.io.*;
import java.util.*;
  
public class R1566B
{
    static int MAX(int x, int y){return ((x > y) ? x : y );}
    static int MIN(int x, int y){return ((x < y) ? x : y );}
    static int sum(int[] A){int sum = 0;for(int i=0; i<A.length; i++)sum+=A[i];return (sum);}
    static int gcd(int a, int b){if (b == 0)return a;return gcd(b, a % b);}
    static boolean isPowerOfTwo(int n){if(n==0)return false;return (int)(Math.ceil((Math.log(n) / Math.log(2)))) ==(int)(Math.floor(((Math.log(n) / Math.log(2)))));}
    static int countDigits(long l)
    {
            if (l >= 1000000000000000000L) return 19;
            if (l >= 100000000000000000L) return 18;
            if (l >= 10000000000000000L) return 17;
            if (l >= 1000000000000000L) return 16;
            if (l >= 100000000000000L) return 15;
            if (l >= 10000000000000L) return 14;
            if (l >= 1000000000000L) return 13;
            if (l >= 100000000000L) return 12;
            if (l >= 10000000000L) return 11;
            if (l >= 1000000000L) return 10;
            if (l >= 100000000L) return 9;
            if (l >= 10000000L) return 8;
            if (l >= 1000000L) return 7;
            if (l >= 100000L) return 6;
            if (l >= 10000L) return 5;
            if (l >= 1000L) return 4;
            if (l >= 100L) return 3;
            if (l >= 10L) return 2;
            return 1;
    }
    static void sort2D(int[][] Arr){Arrays.sort(Arr, (a, b) -> Double.compare(a[0], b[0]));}
    public static void main(String[] args) throws IOException
    {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        if (System.getProperty("ONLINE_JUDGE") == null)
        {
            try 
            {
                in = new BufferedReader(new FileReader("input.txt"));
                System.setOut(new PrintStream(new FileOutputStream("output.txt")));
            }
            catch (Exception e) {
                //System.out.println("Exception");
            }
        }
        int T = Integer.parseInt(in.readLine());
        while( T -- > 0)
        {
            String lines = in.readLine();
            int count = 0;
            int temp = 0;
            int N = lines.length();
            for(int i = 0; i < N; i++)
            {
                char temp1 = lines.charAt(i);
                if(temp1 == '0')
                    count = 1;
                else if(count == 1)
                {
                    count = 0;
                    temp++;
                }
            }
            if(lines.charAt(N-1) == '0')
                temp++;
            System.out.println(MIN(2,temp));
        }
    }
}