Count divisors of product of array elements

Given an array with N elements, the task is to find the count of factors of a number X which is product of all array elements.
Example:
Input : 2 4 6
Output : 10 2 * 4 * 6 = 48,the factors of 48 are 1, 2,
3, 4, 6, 8, 12, 16, 24, 48 whose count is 10.

Solution:
JAVA
import java.util.*;
import java.lang.*;
import java.io.*;

class Solution {
    static List<Integer> ll = new ArrayList<>();
    static  final int MAX = 101;
    static boolean[] isP = new boolean[MAX];
    static Scanner sc = new Scanner(System.in);
public static void main (String[] args) {
//code
init();
int n = sc.nextInt();

while(n-- > 0) solve();
}

public static void solve(){
    int n = sc.nextInt();
    if(n>0){
    int[] arr = new int[n];
    long ans = 1;
    int flag =0;
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    for(int i = 0; i < n; i++){
        arr[i] = sc.nextInt();
        if(arr[i]>=1){
        flag=1;
        countFact(arr[i], map);
}
    }
    for(int key : map.keySet()){
        ans *= (map.get(key) + 1);
    }
    if(flag==1){
    System.out.println(ans);}
}
   
}
static void countFact(int n, Map<Integer, Integer> map){
    int i = 0;
    while(n != 1){
        while(n % ll.get(i) == 0){
            n = n / ll.get(i);
            map.put(ll.get(i),
                map.getOrDefault(ll.get(i), 0) + 1);
        }
        i++;
    }
}
static void init(){
    Arrays.fill(isP, true);
    isP[0] = isP[1] = false;
    for(int i = 2; i * i < MAX; i++){
        if(isP[i]){
            for(int j = i * i; j < MAX; j += i){
                isP[j] = false;
            }
        }
    }
    for(int i = 2; i < MAX; i++){
        if(isP[i]) ll.add(i);
    }
}
}