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You are given two positive integer arrays `nums`

and `numsDivide`

. You can delete any number of elements from `nums`

.

Return *the minimum number of deletions such that the smallest element in *

`nums`

`numsDivide`

. If this is not possible, return `-1`

.Note that an integer `x`

divides `y`

if `y % x == 0`

.

**Example 1:**

```
Input: nums = [2,3,2,4,3], numsDivide = [9,6,9,3,15]
Output: 2
Explanation:
The smallest element in [2,3,2,4,3] is 2, which does not divide all the elements of numsDivide.
We use 2 deletions to delete the elements in nums that are equal to 2 which makes nums = [3,4,3].
The smallest element in [3,4,3] is 3, which divides all the elements of numsDivide.
It can be shown that 2 is the minimum number of deletions needed.
```

**Example 2:**

```
Input: nums = [4,3,6], numsDivide = [8,2,6,10]
Output: -1
Explanation:
We want the smallest element in nums to divide all the elements of numsDivide.
There is no way to delete elements from nums to allow this.
```

**Constraints:**

`1 <= nums.length, numsDivide.length <= 10`

^{5}`1 <= nums[i], numsDivide[i] <= 10`

^{9}

```
public int minOperations(int[] A, int[] numsDivide) {
int g = numsDivide[0], tmp;
for (int a : numsDivide) {
while (a > 0) { // g = gcd(g, a)
tmp = g % a;
g = a;
a = tmp;
}
}
Arrays.sort(A);
for (int i = 0; i < A.length && A[i] <= g; ++i)
if (g % A[i] == 0)
return i;
return -1;
}
```

```
int minOperations(vector<int>& A, vector<int>& numsDivide) {
int g = numsDivide[0];
for (int a: numsDivide)
g = gcd(g, a);
sort(A.begin(), A.end());
for (int i = 0; i < A.size() && A[i] <= g; ++i)
if (g % A[i] == 0)
return i;
return -1;
}
```

```
def minOperations(self, A: List[int], numsDivide: List[int]) -> int:
g = reduce(gcd, numsDivide)
for i,a in enumerate(sorted(A)):
if g % a == 0: return i
if a > g: break
return -1
```

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