D. Soldier and Number Game
　　　　　　　　　　　　　　　　　　　　　　　　　　time limit per test3 seconds
　　　　　　　　　　　　　　　　　　　　　　　　　　memory limit per test256 megabytes
　　　　　　　　　　　　　　　　　　　　　　　　　　inputstandard input
　　　　　　　　　　　　　　　　　　　　　　　　　　outputstandard output
Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to the second soldier. Then the second one tries to make maximum possible number of rounds. Each round consists of choosing a positive integer x > 1, such that n is divisible by x and replacing n with n / x. When n becomes equal to 1 and there is no more possible valid moves the game is over and the score of the second soldier is equal to the number of rounds he performed.

To make the game more interesting, first soldier chooses n of form a! / b! for some positive integer a and b (a ≥ b). Here by k! we denote the factorial of k that is defined as a product of all positive integers not large than k.

What is the maximum possible score of the second soldier?

Input
First line of input consists of single integer t (1 ≤ t ≤ 1 000 000) denoting number of games soldiers play.

Then follow t lines, each contains pair of integers a and b (1 ≤ b ≤ a ≤ 5 000 000) defining the value of n for a game.

Output
For each game output a maximum score that the second soldier can get.

Sample test(s)
input
2
3 1
6 3
output
2
5


　　



/*
    dp求解 n！的质因子个数。 
    1.首先利用素数筛选法求出素数的集合
    2.dp[i] = dp[i/prime[j]]+1, i%prime[j]==0; prime[j]是第一个能整除i的数 
*/
#include<iostream> 
#include<cstring>
#include<cstdio>
#include<algorithm>
#define N 5000005
using namespace std;

int prime[N];
bool isprime[N];
int dp[N];
void init(){
    int top = 0;
    for(int i=2; i<N; ++i){
        if(!isprime[i])
            prime[top++] = i;
        for(int j=0; j<top && i*prime[j]<N; ++j){//素数筛选 
            isprime[i*prime[j]] = true;
            if(i%prime[j] == 0)
                break;
        }
    }
    for(int i=2; i<N; ++i){
        if(!isprime[i])
            dp[i] = 1;
        else{
            for(int j=0; j<top; ++j)
                if(i%prime[j]==0){
                    dp[i] = dp[i/prime[j]] + 1;
                    break;
                }
        }
    }
    for(int i=2; i<N; ++i)//前n个数的质因数个数的和 
        dp[i]+=dp[i-1];
}

int main(){
    int t;
    scanf("%d", &t);
    init();
    while(t--){
        int a, b;
        scanf("%d%d", &a, &b);
        printf("%d\n", dp[a] - dp[b]);
    }
    return 0; 
}