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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

#define mod 1e9+7
#define ll long long int
#define epsilon 1e-7

int main(void){
    int q;
    scanf("%d", &q);
    int l[q], r[q];
    for(int i = 0;i < q;i++){
        scanf("%d %d", &l[i], &r[i]);
    }
    int c[100001];
    int prime[65] ={2,3,5,7,11,13,17,19,23,29,
    31,37,41,43,47,53,59,61,67,71,
    73,79,83,89,97,101,103,107,109,113,
    127,131,137,139,149,151,157,163,167,173,
    179,181,191,193,197,199,211,223,227,229,
    233,239,241,251,257,263,269,271,277,281,
    283,293,307,311,313};
    c[0] = 0;
    c[1] = 0;
    for(int i = 2;i < 100001;i++){
        int j = 0;
        while(j < 65){
            if(i < prime[j]*prime[j]){
                c[i] = 1;
                break;
            }
            if(i%prime[j]==0){
                c[i] = 0;
                break;
            }
            j++;
        }
    }
    int cc[100001];
    int count = 0;
    for(int i = 0;i < 100001;i++){
        if(c[i] == 1 && c[(i+1)/2] == 1){
            count++;
        }
        cc[i] = count;
    }
    for(int i = 0;i < q;i++){
        printf("%d\n", cc[r[i]]-cc[l[i]-1]);
    }
    return 0;
}