Clever use of Prime sieve 2

Description
Recently, the bear started studying data structures and faced the following problem.

You are given a sequence of integers x1, x2, ..., xn of length n and M queries, each of them was characterized by S Li, RI. Let's introduce F (p) to represent the number of such indexes K, that XK was divisible by P. The answer to the query Li, ri are the sum:, where S (Li, RI) is a set of prime numbers from segment [Li, RI] (both borders is included in the segment).

Help the Bear cope with the problem.

Input
The first line contains integer n (1≤n≤106). The second line contains n integers x1, x2, ..., xn (2≤xi≤107). The numbers is not necessarily distinct.

The third line contains integer m (1≤m≤50000). Each of the following m lines contains a pair of space-separated integers, Li and ri (2≤li≤ri≤2 109)-the numbers th At characterize, the current query.

Output
Print m integers-the answers to the queries on the order the queries appear in the input.

Sample Input
Input
6
5 5 7 10 14 15
3
2 11
3 12
4 4
Output
9
7
0
Input
7
2 3 5 7 11 4 8
2
8 10
2 123
Output
0
7
Hint
Consider the first sample. Overall, the first sample has 3 queries.

The first query L = 2, R = comes. You need to count F (2) + F (3) + F (5) + F (7) + f (11) = 2 + 1 + 4 + 2 + 0 = 9.
The second query comes L = 3, R = 12. You need to count F (3) + F (5) + F (7) + f (11) = 1 + 4 + 2 + 0 = 7.
The third query comes L = 4, R = 4. As this interval have no prime numbers, then the sum equals 0.

At the beginning of the thought might use a tree-like array or something, the result is not as violent as others

 #include <cstdio> #include <iostream> using namespace std; const int max=10000010;
int n,cnt[max],vis[max],is[max],m,a,b,t;
            void Init () {for (int i=2;i<max;i++) {if (!is[i]) {cnt[i]+=vis[i];
        for (int j=i+i;j<max;j+=i) IS[J]=1,CNT[I]+=VIS[J];
}} for (int i=2;i<max;i++) cnt[i]+=cnt[i-1];
    } int main () {scanf ("%d", &n);
        for (int i=0;i<n;i++) {scanf ("%d", &t);
    vis[t]++;
    } init ();
    scanf ("%d", &m);
        while (m--) {scanf ("%d%d", &a,&b);
        if (a>max-1) a=max-1;
       if (b>max-1) b=max-1;
        cout<< "----" <<cnt[b]<< "" <<cnt[a-1]<<endl;
    printf ("%d\n", Cnt[b]-cnt[a-1]);
} return 0; }