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.util.*;

public class Test {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		sc.nextLine();//dump
		while(t--!=0) {
			String s=sc.nextLine();
			int zero=0,one=0;//count
			for(int i=0;i<s.length();i++) {
				if(s.charAt(i)=='0')
					zero++;
				else one++;
			}
//			System.out.println(zero+" "+one);
			if(one==0) { //only zeros
				System.out.println(1);
			}else if(zero==0) { //only ones
				System.out.println(0);
			}else {
				int i=0;
				boolean flag=false;
				if(s.charAt(0)!='0' && s.charAt(s.length()-1)!='0') {
					while(i<s.length() && s.charAt(i)=='1') {
						i++;
					}
					int j=s.length()-1;
					while(j>=i && s.charAt(i)=='1') {
						j--;
					}
					for(;i<=j;i++) {
						if(s.charAt(i)=='1') {
							flag=true;
							break;
						}
					}
				}
				if(!flag) { 
					if(zero>1)
						System.out.println(2);
					else
						System.out.println(1);
				}else {
					System.out.println(1);
				}
			}
		}
	}
}
