Problem:
Score : 400 points

Problem Statement

We say that a odd number N is similar to 2017 when both N and (N+1)/2 are prime.

You are given Q queries.

In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.

Constraints

1≤Q≤10^5

1≤l_i≤r_i≤10^5

l_i and r_i are odd.

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q
l_1 r_1
:
l_Q r_Q

Output

Print Q lines. The i-th line (1≤i≤Q) should contain the response to the i-th query.

Sample Input 1

1
3 7

Sample Output 1

2

3 is similar to 2017, since both 3 and (3+1)/2=2 are prime.

5 is similar to 2017, since both 5 and (5+1)/2=3 are prime.

7 is not similar to 2017, since (7+1)/2=4 is not prime, although 7 is prime.

Thus, the response to the first query should be 2.

Sample Input 2

4
13 13
7 11
7 11
2017 2017

Sample Output 2

1
0
0
1

Note that 2017 is also similar to 2017.

Sample Input 3

6
1 53
13 91
37 55
19 51
73 91
13 49

Sample Output 3

4
4
1
1
1
2

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<math.h>
#include<vector>
#include<algorithm>
#include<numeric>
#include<map>
#include<queue>
#include<deque>
#include<set>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
typedef long long ll;
typedef vector<int> vec;
typedef vector<vector<int>> mat;

int main(){
    //S[i]:i-2までに含まれる2017に似た奇数の個数
    int S[100010];
    S[3]=S[1]=0;

    int i=3,N,M,root_N,root_M,k,l;
    bool prime_N,prime_M;
    while(i<=100000){
        N=i,M=(i+1)/2,root_N=sqrt(N)/1;root_M=sqrt(M)/1;
        
        k=3;prime_N=true;
        while(k<=root_N){
            if(N%k==0){
                prime_N=false;
                break;
            }
            k+=2;
        }

        l=2;prime_M=true;
        while(l<=root_M){
            if(M%l==0){
                prime_M=false;
                break;
            }
            l++;
        }

        if(prime_N && prime_M)S[i+2]=S[i]+1;
        else S[i+2]=S[i];
        i+=2;
    }

    int Q;cin>>Q;
    int s,t;
    for(int i=0;i<Q;i++){
        cin>>s>>t;
        cout<<S[t+2]-S[s]<<endl;
    }
    return 0;
}