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# Digit multiplier

Given a positive integer **N**, find the smallest number **S** such that the product of all the digits of **S** is equal to the number **N**.

**Input:**\
The first line of input contains an integer **T** denoting the number of test cases. Then **T** test cases follow. The first line of each test case contains a positive integer N**.**

**Output:**\
Print out the number **S**. If there exists no such number, then print **-1**.

**Constraints:**\
1 <= T <= 200\
1 <= N <= 109

**Examples:** \
**Input:**\
2\
100\
26\
**Output:**\
455\
-1

```java
class Solution {
    // 1st prefernce is reducing the digits
    // 2nd prefernce is forming min digit among same digits set
    public long minDigit(int n) {
        if(n==1)
            return 1;
        // For getting minimum number of digits
        // We take bigger factors first(9 8 7 ...2)
        // Double digits factors will be covered by single digits
        List<Integer> digits = new ArrayList<>();
        int factor = 9;
        // 1st Prefernce
        while (n > 1 && factor > 1) {
            while (n % factor == 0) {
                n = n / factor;
                digits.add(factor);
            }
            factor--;
        }
        // Edge case -> PrimeNumber
        if (n != 1)
            return -1;
        // 2nd prefernce
        // Now our list contains digits in decreasing order
        // Min ans will be reverse of this
        long ans = 0;
        for (int i = digits.size() - 1; i >= 0; i--)
            ans = ans * 10 + digits.get(i);
        return ans;
    }
}
```