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 static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;

public class A{
    public static void main(String[] args) throws IOException {
        FastScanner s = new FastScanner();
            int t = s.nextInt();
//        int t = 1;
        while (t-- > 0) {
            String g=s.next();
            if(g.charAt(g.length()-1)=='0')g+=1;
            else g+=0;
            char[] a = g.toCharArray();
            long ans=0;
            for(int i=1;i<a.length;i++){
                if(a[i]==a[i-1])continue;
                else{
                    if(a[i-1]=='0')ans++;
                }
            }
            if(ans==0){
                if(a[0]=='1') out.println(0);
                else out.println(1);
                continue;
            }
            out.println(min(ans,2));
        }
    }
    static long gcd(long a, long b) throws IOException {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static int gcd(int a, int b) throws IOException {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static void sortr(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a) l.add(i);
        Collections.sort(l, Collections.reverseOrder());
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }

    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a) l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }

    static void sort(long[] a) {
        ArrayList<Long> l = new ArrayList<>();
        for (long i : a) l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }
}
