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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include<stdio.h>
typedef struct tag_q
{
    int l;
    int r;
}query;

int main(void){
    int number_flag[10000];
    int q_num;
    scanf("%d",&q_num);
    query Query[q_num];
    for(int i = 0;i < q_num;i++){
        scanf("%d %d",&(Query[i].l),&(Query[i].r));
    }

    //all flag set 1(prime)
    for(int i = 0;i < 10000;i++){
        number_flag[i] = 1;
    }

    number_flag[0] = 0;//1 is not prime

    //check prime(flag1 = prime)

    for(int i = 1;i < 10000;i++){
        if(number_flag[i] == 1){
            for(int j = 2;((i + 1) * j - 1) < 10000;j++){
                number_flag[(i + 1) * j - 1] = 0;
            }
        }
    }

    //2017_Like Check(flag1 = 2017_Like)
    //all flag set 0(not Like)
    int Like_Check_num[10000];

    for(int i = 0;i < 10000;i++){
        Like_Check_num[i] = 0;
    }

    for(int i = 2;i < 10000;i++){
        if(number_flag[i] == 1 && number_flag[((i + 2) / 2 )- 1] == 1){
            Like_Check_num[i] = 1;
            //printf("%d\n",i + 1);
        }
    }

    int number_cusum[10001];
    number_cusum[0] = 0;

    for(int i = 1;i < 10001;i++){
        number_cusum[i] = number_cusum[i - 1] + Like_Check_num[i - 1];
        //printf("%d",number_cusum[i]);
    }
    
    //res query
    for(int i = 0;i < q_num;i++){
        //printf("%d - %d\n",Query[i].r + 1,Query[i].l);
        printf("%d\n",number_cusum[Query[i].r] - number_cusum[Query[i].l - 1]);
    }
    return 0;
}