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 <cstring>
#include <cstdio>
#include<utility>
#include<algorithm>
#include<vector>
#include<cmath>
#include<map>
#include<bitset>
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
typedef pair<double,double>P_D;
#define PI 3.1415926535897932384626433832795
#define rep(i, n) for (int i = 0; i < (int)n; i++)
#define reprev(i, n) for (int i = (int)n - 1; i >= 0; i--)
#define ALL(a)  (a).begin(),(a).end()
#define MAX 100100
#define MAX_N 200100
#define SENTINEL 2000000000
#define MAX_B 1001001
#define NIL -1
const int MOD = 1e9+7;
const int INF = 1 << 29;
const ll INFTY=1000000000000000000LL;
int main(){
    //まず素数表を完成させる->エラストテネスのふるい
    bitset<MAX_B>is_prime(0);
    for(int i=2;i<MAX;i++)is_prime.set(i);
    for(int i=2;i<MAX;++i){
        if(is_prime[i]){
            for(int j=i*2;j<MAX;j+=i){
                is_prime.set(j,0);
            }
        }
    }
    //2017-like
    bool a[MAX];
    for(int i=1;i<MAX;i+=2){
        if(is_prime[i] && is_prime[(i+1)/2]){
            a[i]=1;
        }
    }
    //累積和
    int sum[MAX];
    sum[1]=0;
    for(int i=1;i<MAX-1;i++){
        sum[i+1]=sum[i]+a[i];
    }
    //あとは問題の処理
    int q,l,r;
    cin>>q;
    rep(i,q){
        cin>>l>>r;
        cout<<sum[r+1]-sum[l]<<endl;
    }
}