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.
//package com.codeforces;

import java.io.IOException;
import java.util.Scanner;

public class Solution {

    public static void main(String[] args) throws IOException {
//        File file = new File("/home/nobroker/Desktop/input.txt");
        Scanner sc = new Scanner(System.in);
//        FileWriter fileWriter = new FileWriter("/home/nobroker/Desktop/output.txt");
//        PrintWriter printWriter = new PrintWriter(fileWriter);
        int t = Integer.parseInt(sc.nextLine());
        for (int i = 1; i <= t; ++i) {
            String str = sc.nextLine();
            int zCount = 0, oCount = 0;
            for (int j = 0 ; j < str.length() ; j++) {
                if (str.charAt(j) == '0') {
                    zCount++;
                } else {
                    oCount++;
                }
            }
            int ans = 1;
            if (zCount == 0) {
              ans = 0;
            } else if (oCount == 0) {
                ans = 1;
            } else {
                int pCount = 0;
                for (int j = 0 ; j < str.length() - 1 ; j++) {
                    if (str.charAt(j) == '0' && str.charAt(j + 1) == '1') {
                        pCount++;
                    } else {
                        if (pCount > 0 && str.charAt(j) == '0') {
                            ans = 2;
                            break;
                        }
                    }
                }
                if (pCount > 0 && str.charAt(str.length() - 1) == '0') {
                    ans = 2;
                }
            }
            System.out.println(ans);
        }
//        printWriter.close();
    }


}
