CodeForces - 546D Soldier and Number Game (素数打表)

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 ：让你求 a+1 到 b的数字中有多少个素数  

如 6    3      4 （2*2） 5   6(2*3)  有5个

解 素数打表 

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
#define inf 0x3f3f3f3f
#define ll long long
#define maxx 5000005
using namespace std;
int ans[maxx];
bool v[maxx];//是否为素数
int t,a,b;
void pp()//在这边 ans数组为 每个数字分解的素因子个数
{
    for(int i=2;i<maxx;i++)
    {
        if(v[i]==0)//如果他为素数
        {
            for(int j=i;j<maxx; j+=i)//这个素数的倍数
            {
                int u=j; 
                while(u%i==0)  //计算倍数中有几个该素因子
                {
                    u/=i;
                    ans[j]++; 
                }
                v[j]=1; //标记为非素数
            }
        }
    }
}
int main()
{
    memset(ans,0,sizeof(ans));
    memset(v,0,sizeof(v));
    pp();
    for(int i=1;i<maxx;i++)//计算和
        ans[i]+=ans[i-1];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&a,&b);
        printf("%d\n",ans[a]-ans[b]);
    }
}