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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define ll long long
using namespace std;

bool isPrime(int x){
    int i;
    if(x < 2)return 0;
    else if(x == 2) return 1;
    if(x%2 == 0) return 0;
    for(i = 3; i*i <= x; i += 2) if(x%i == 0) return 0;
    return 1;
}

int main(){
    ll int n,a[100001],b[100001],c[100001],l[100001],r[100001];
    c[0]=0;
    for (int i=1;i<100001;i++){
        if(isPrime(i)){
            a[i]=1;
            if(a[(i+1)/2]==1){
                b[i]=1;
            }else{
                b[i]=0;
            }
        }else{
            a[i]=0;
            b[i]=0;
        }
        c[i]=c[i-1]+b[i];
    }
    cin >> n;
    for (int i=0;i<n;i++){
        cin >> l[i] >> r[i];
        cout << c[r[i]]-c[l[i]-1] << endl;
    }
    
}