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 <stdio.h>
int isPrime(int n){
    if(n == 1){
        return 0;
    }
    if(n == 2){
        return 1;
    }
    if(n == 3){
        return 1;
    }
    if(n == 5){
        return 1;
    }
    if(n%2 == 0){
        return 0;
    }
    if(n%3 == 0){
        return 0;
    }
    if(n%5 == 0){
        return 0;
    }
    for(int i = 3; i*i <= n; i = i + 2){
        if(n % i == 0){
            return 0;}
    }
    return 1;
}

int main(void){
    int q;
    scanf("%d", &q);
    int l[500000], r[500000];
    int i, j;
    int cnt;
    for(i = 0; i < q; i++){
        scanf("%d%d", &l[i], &r[i]);
    }
    int cumsum[500000];
    cumsum[1] = 0;
    for(i = 3; i <= 500000; i = i + 2){
        if((isPrime(i) == 1) && (isPrime( (i + 1)/ 2) ) ){
            cumsum[i] = cumsum[i-2] + 1;
        }else
        {
            cumsum[i] = cumsum[i-2];
        }
    }
    for(i = 0; i < q; i++){
        if(l[i] == 1){
            printf("%d\n", cumsum[r[i]]);
        }else{
            printf("%d\n", cumsum[r[i]] - cumsum[l[i] - 2]);
        }
    }
    return 0;
}