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 <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

#define REP(i, N) for(int (i) = 0; (i) < (N); i++)
#define MAX 1 << 20

bool isPrime(int n){
    if(n <= 1) return false;
    if(n == 2) return true;
    for(int i = 2; i*i <= n; i++)
        if(n % i == 0) return false;

    return true;
}

int main(){
    int n, a[50001], ans[100000];
    scanf("%d", &n);

    a[0] = 0;
    REP(i, 50000){
        int p = i*2+1;
        a[i+1] = a[i] + (isPrime(p) && isPrime((p+1)/2) ? 1 : 0);
    }

    int li, ri; 
    REP(i, n){
        scanf("%d %d", &li, &ri);
        ri = (ri+1)/2;
        li = (li+1)/2;
        ans[i] = a[ri] - a[li-1];
    }
    REP(i, n)
        printf("%d\n", ans[i]);
    return 0;
}