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.*;
 
public class B {
	static int xor[] = new int[1000000];
	
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int t = sc.nextInt();
		for(int tt = 0; tt < t; tt++) {
			String s = sc.next();
			
			int d = 0;
			for(int i = 0 ; i < s.length() ; i++) {
				if(i == 0 && s.charAt(i) == '0')d++;
				else {
					if( s.charAt(i) == '0' && s.charAt(i) != s.charAt(i-1))d++;
				}
			}
			if(d >= 2)System.out.println(2);
			else {
				if(d == 1)System.out.println(1);
				else System.out.println(0);
			}
		}
	}

	public 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);
	}

  
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }
 
    public static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
    public static long pow(long base, long exp, long MOD) {
        base %= MOD;
        long ret = 1;
        while (exp > 0) {
            if ((exp & 1) == 1) ret = ret * base % MOD;
            base = base * base % MOD;
            exp >>= 1;
        }
        return ret;
    }
 
 
    public static long max(long... ar) {
        long ret = ar[0];
        for (long itr : ar) ret = Math.max(ret, itr);
        return ret;
    }
 
    public static int max(int... ar) {
        int ret = ar[0];
        for (int itr : ar) ret = Math.max(ret, itr);
        return ret;
    }
 
    public static long min(long... ar) {
        long ret = ar[0];
        for (long itr : ar) ret = Math.min(ret, itr);
        return ret;
    }
 
    public static int min(int... ar) {
        int ret = ar[0];
        for (int itr : ar) ret = Math.min(ret, itr);
        return ret;
    }
 
 
    public static long sum(long... ar) {
        long sum = 0;
        for (long itr : ar) sum += itr;
        return sum;
    }
 
    public static long sum(int... ar) {
        long sum = 0;
        for (int itr : ar) sum += itr;
        return sum;
    }
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		
		public String next() {
			while (!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());
		}
		
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
	}
 
}


 
 
 
 
 
 
 
 
