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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;
import java.util.Vector;



public class MINMEX {
    public static void main(String [] args) {
        FastReader fs = new FastReader();
        int t = fs.nextInt();
        for (int i = 0; i < t; i++) {
            String str = fs.next();
            solve(str);
        }
    }

    static void solve(String str) {
        int res = 0;
        char current = str.charAt(0);
        if (current == '0') res++; 
        for (int i = 1; i < str.length(); i++) {
            char a = str.charAt(i);
            if (current != a) {
                if (a == '0') res+=1;         
                current = a;
            }

            if (res >= 2) 
            {
                System.out.println(res);
                return;
            }
        }
        System.out.println(res);
    }

    static int gcd (int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a%b);
    }

    static void pa(int arr[]) {
        System.out.println(Arrays.toString(arr));
    }
    static void pa(String arr[]) {
        System.out.println(Arrays.toString(arr));
    }

    static int[] sieve(int n) {
        int prime[] = new int[n];
        Arrays.fill(prime, 1);
        for (int i = 2; i < n; i++) {
            if (prime[i] == 1) {
                for (int j = i*i; i <= n; j+=i) {
                    prime[i] = 0;
                }
            }
        }
        return prime;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader()
        {
            br = new BufferedReader(
                new InputStreamReader(System.in));
        }
        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() { return Integer.parseInt(next()); }
 
        long nextLong() { return Long.parseLong(next()); }
 
        double nextDouble() { return Double.parseDouble(next()); }
 
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}
