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.*;
import java.util.*;

public class Main {
	static MyScanner sc = new MyScanner();
	static Output out=new Output();
	public static void main(String[] args) {
		int t=sc.nextInt();
		for(int i=0;i<t;i++) {
			solve();
		}
		out.flush();
	}
	
	public static void solve() {
		String s=sc.next();
		int c0=0,c1=0;
		int l=s.length();
		for(int i=0;i<l;i++) {
			if(s.charAt(i)=='0')
				c0++;
			else
				c1++;
		}
		if(c0==l)
			out.println(1);
		else if(c1==l)
			out.println(0);
		else {
			int cc0=0;
			for(int i=0;i<l;i++) {
				if(s.charAt(i)=='0') {
					for(int j=i;j<l;j++) {
						if(s.charAt(j)=='0')
							cc0++;
						else
							break;
					}
					break;
				}
			}
			if(cc0==c0)
				out.println(1);
			else
				out.println(2);
		}
	}	
}























class MyScanner {

	BufferedReader br;
	StringTokenizer st;

	public MyScanner() {
		br = new BufferedReader(new InputStreamReader(System.in));
	}

	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;
	}

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

	public long[] nextLongArr(int n) {
		long a[] = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}

	public double[] nextDoubleArr(int n) {
		double a[] = new double[n];
		for (int i = 0; i < n; i++)
			a[i] = nextDouble();
		return a;
	}

	public String[] nextArr(int n) {
		String a[] = new String[n];
		for (int i = 0; i < n; i++)
			a[i] = next();
		return a;
	}

	public String[] nextLineArr(int n) {
		String a[] = new String[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLine();
		return a;
	}
}
class Output{
	StringBuilder s=new StringBuilder();
	public void println() {
		s.append("\n");
	}
	public void println(Object x) {
		s.append(x+"\n");
	}
	
	public void print(Object x) {
		s.append(x+"");
	}
	
	public void flush() {
		System.out.println(s);
		s=new StringBuilder();
	}
}