In many programming competitions, we are asked to find (or count the number of) Prime Factors of an integer i. This is boring. This time, let’s count the number of Non-Prime Factors of an integer i

, denoted as NPF(i).

For example, integer 100
has the following nine factors: {1,2⎯⎯,4,5⎯⎯,10,20,25,50,100}. The two which are underlined are prime factors of 100 and the rest are non-prime factors. Therefore, NPF(100) = 7

.
Input

The first line contains an integer Q
(1≤Q≤3⋅106) denoting the number of queries. Each of the next Q lines contains one integer i (2≤i≤2⋅106

).
Output

For each query i

, print the value of NPF(i).
Warning

The I/O files are large. Please use fast I/O methods.
Sample Input 1 Sample Output 1

4
100
13
12
2018

7
1
4
2

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define maxx 2e6+5

int vis[2000002];
int ans[2000002];
void f()
{
    vis[1]=1;
    for(int i=2; i<=sqrt(maxx); ++i)
    {
        if(!vis[i])
        {
            for(int j=i*i; j<=2000002; j+=i)
            {
                vis[j]=1;
            }
        }
    }
    for(int i = 1; i <= 2000000; ++i)
    {
        int rt = 2000000/i;
        for(int j = i; j <= rt; ++j)
        {
            if(vis[i])
            {
                ans[i*j]++;
            }
            if(vis[j] && i!=j)
            {
                ans[i*j]++;
            }
        }
    }
}

int main()
{
    int t;
    f();
    scanf("%d", &t);
    while(t--)
    {
        int n;
        scanf("%d", &n);
        printf("%d\n", ans[n]);
    }
    return 0;
}