You are given an integer N. Find the digits in this number that exactly divide N (division that leaves 0 as remainder) and display their count. For N=24, there are 2 digits (2 & 4). Both of these digits exactly divide 24. So our answer is 2.

**Note**

- If the same number is repeated twice at different positions, it should be counted twice, e.g., For N=122, 2 divides 122 exactly and occurs at ones’ and tens’ position. So for this case, our answer is 3.
- Division by 0 is undefined.

**Input Format**

The first line contains T (the number of test cases), followed by T lines (each containing an integer N).

**Constraints**

1≤T≤15

0<N<1010

**Output Format**

For each test case, display the count of digits in N that exactly divide N in a separate line.

**Sample Input**

```
2
12
1012
```

**Sample Output**

```
2
3
```

**Explanation**

2 digits in the number 12 divide it exactly. The first digit, 1, divides it exactly in twelve parts, and the second digit, 2 divides 12 equally in six parts.

1 divides 1012 exactly (and it occurs twice), and 2 also divides it exactly. Division by 0 is undefined, therefore it will not be counted.

class Find_Digits_Hackerrank

{

static void Main(string[] args)

{

int T = Convert.ToInt16(Console.ReadLine());

int count = 0;

for (int i = 0; i < T; ++i)

{

count = 0;

string N = Console.ReadLine();

foreach (char ch in N)

{

if ((ch != ‘0’) && (Convert.ToInt64(N) % Convert.ToInt16(ch – ‘0’) == 0))

{

++count;

}

}

Console.WriteLine(count);

}

}

}

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