Codeforces Round #304 (Div. 2) D. Soldier and Number Game 素数打表+质因数分解

D. Soldier and Number Game

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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.

Examples

Input

2
3 1
6 3

Output

2
5
题意：求[b+1,a]区间内的所有数质因数分解的得到的的质数个数；
思路：素数打表+质因数分解求前缀和，这题nsqrt（n）不知道能不能过；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define esp 1e-10
const int N=1e6+10,M=1e6+10,mod=1e9+7,inf=1e9+10;
const int MAXN=5000001;
int prime[MAXN];
bool vis[MAXN];
int sum[MAXN];
int Prime(int n)
{
    int cnt=0;
    memset(vis,0,sizeof(vis));
    for(int i=2;i<n;i++)
    {
        if(!vis[i])
        prime[cnt++]=i;
        for(int j=0;j<cnt&&i*prime[j]<n;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)
            break;
        }
    }
    return cnt;
}
int main()
{
    int x,y,z,i,t;
    int cnt=Prime(MAXN);
    for(i=1;i<MAXN;i++)
    {
        int flag=i;
        for(t=0;t<cnt;t++)
        {
            if(flag==1)
            break;
            if(!vis[flag])
            {
                sum[i]++;
                break;
            }
            while(flag%prime[t]==0)
            {
                flag/=prime[t];
                sum[i]++;
            }
        }
        sum[i]+=sum[i-1];
    }
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        printf("%d
",sum[a]-sum[b]);
    }
    return 0;
}