No-Zero integer is a positive integer that does not contain any 0 in its decimal representation.

Given an integer n, return a list of two integers [A, B] where:


	A and B are No-Zero integers.
	A + B = n


The test cases are generated so that there is at least one valid solution. If there are many valid solutions you can return any of them.

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Example 1:

Input: n = 2
Output: [1,1]
Explanation: A = 1, B = 1. A + B = n and both A and B do not contain any 0 in their decimal representation.


Example 2:

Input: n = 11
Output: [2,9]


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Constraints:


	2 &lt;= n &lt;= 104


class Solution:
    def getNoZeroIntegers(self, n: int) -> List[int]:
        for i in range(1,n//2+1):
            first = str(i)
            second = str(n-i)
            if "0" not in first and "0" not in second:
                return [i, n-i]
class Solution {
    public int[] getNoZeroIntegers(int n) {
        int B;
        for (int A = 1; A < n; ++A) {
            B = n - A;
            if (!(A + "").contains("0") && !(B + "").contains("0"))
            return new int[] {A, B};
    }
        return new int[]{};
}
}
class Solution {
public:
    int has0(int x)
{
    while (x){
        if (x % 10 == 0)
          return 1;
        x /= 10;
    }
    return 0;
}
    vector<int> getNoZeroIntegers(int n) {
        for(int i=1;i<=n;i++){
            if(has0(i)==false && has0(n-i)==false){
                return {i,n-i};
            }
        }
        return {1,1};
    }
};
var getNoZeroIntegers = function(n) {
   for(let i=1;i<=n;i++){
       if(!haveZero(i) && !haveZero(n-i)){
           return [i,n-i]
       }
   }
};

const haveZero = (n) =>{
    let copy = n;
    while(copy>0){
        if(copy%10===0){
            return true
        }
        copy=Math.floor(copy/10)
    }
    return false
}
Convert Integer to the Sum of Two No-Zero Integers