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.

// Working program using Reader Class
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.lang.Math;

public class Main {
	public static void solve(String s) {

		// reference link to submit your solution once system testing is done.
		// link: https://codeforces.com/contest/1566/problem/B
		int mex = 0;
		int zeros = 0;
		int ones = 0;
		int n = s.length();
		boolean two_substrings_present = false;

		// count 0 and 1's occurances
		for (int i = 0; i < n; i++) {
			if (s.charAt(i) == '0')
				zeros++;
			if (s.charAt(i) == '1')
				ones++;
		}

		// if all zeros, then answer is 1.
		if (zeros == n)
			mex = 1;

		// if all ones, then answer is 0.
		else if (ones == n)
			mex = 0;

		// if two substrings of 0, then answer is 2.
		else {
			// get first and last occurance of '0',
			// and then check if '1' occurs between them.

			int first = -1;
			int last = -1;

			// find first occurance
			for (int i = 0; i < n; i++) {
				if (s.charAt(i) == '0') {
					first = i;
					break;
				}
			}

			// find last occurance
			for (int i = n - 1; i > 0; i--) {
				if (s.charAt(i) == '0') {
					last = i;
					break;
				}
			}

			// check if '1' exist between first and last occurance of '0'
			for (int i = first; i < last; i++) {
				if (s.charAt(i) == '1') {
					two_substrings_present = true;
					mex = 2;
					break;
				}
			}

			if (!two_substrings_present)
				mex = 1;
		}

		System.out.println(mex);
		// System.out.println("n= " + n + " s= " + s + " median= " + median);
	}

	public static void main(String[] args) throws IOException {
		// Reader sc = new Reader();
		Scanner sc = new Scanner(System.in);
		int tc = sc.nextInt();

		while (tc-- > 0) {
			String s = sc.next();
			solve(s);
		}
		sc.close();
	}

}