题目链接：Problem - 1611A - Codeforces

Polycarp has an integer nn that doesn't contain the digit 0. He can do the following operation with his number several (possibly zero) times:

• Reverse the prefix of length l (in other words, ll leftmost digits) of nn. So, the leftmost digit is swapped with the ll-th digit from the left, the second digit from the left swapped with (l−1)-th left, etc. For example, if n=123456789 and l=5， then the new value of nn will be 543216789.

Note that for different operations, the values of ll can be different. The number ll can be equal to the length of the number nn — in this case, the whole number nn is reversed.

Polycarp loves even numbers. Therefore, he wants to make his number even. At the same time, Polycarp is very impatient. He wants to do as few operations as possible.

Help Polycarp. Determine the minimum number of operations he needs to perform with the number nn to make it even or determine that this is impossible.

You need to answer tt independent test cases.

Input

The first line contains the number tt (1≤t≤104) — the number of test cases.

Each of the following tt lines contains one integer nn (1≤n<109). It is guaranteed that the given number doesn't contain the digit 0.

Output

Print tt lines. On each line print one integer — the answer to the corresponding test case. If it is impossible to make an even number, print -1.

Example

input

Copy

4
3876
387
4489
3

output

Copy

0
2
1
-1

Note

In the first test case, n=3876, which is already an even number. Polycarp doesn't need to do anything, so the answer is 0.

In the second test case,n=387. Polycarp needs to do 2 operations:

1. Select l=2 and reverse the prefix 387. The number nn becomes 837. This number is odd.
2. Select l=3 and reverse the prefix 837. The number nn becomes 738. This number is even.

It can be shown that 2 is the minimum possible number of operations that Polycarp needs to do with his number to make it even.

In the third test case, n=4489. Polycarp can reverse the whole number (choose a prefix of length l=4). It will become 98449844 and this is an even number.

In the fourth test case, n=3. No matter how hard Polycarp tried, he would not be able to make an even number.

题意：给一个数字，你可以选择任意长度（开始位置只能是第一个）把这串反转，问需要几步才能变成偶数

思路：这题答案只有四种情况，如果没有偶数数字就是-1，如果最后一个是偶数就是0，第一个是偶数是1，否则就是2

#include<bits/stdc++.h>
using namespace std;


int main(){ 
	int t;
	cin >> t;
	while(t--){
		string s;
		cin >> s;
		bool flag = 0;
		for(int i = 0; i < s.length(); i++){
			if(s[i] == '2' || s[i] == '0' || s[i] == '4' || s[i] == '6' || s[i] == '8'){
				flag = 1;
			}
		}
		if(flag == 0){
			cout << -1 << endl;
			continue;
		}
		int len = s.length();
		if(s[len - 1] == '0' || s[len - 1] == '2' || s[len - 1] == '4' || s[len - 1] == '6' || s[len - 1] == '8'){
			cout << 0 << endl;
		}else if((s[0] - '0') % 2 == 0){
			cout << 1 << endl;
		}else{
			cout << 2 << endl;
		}
	} 
	
	return 0;
}