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: Time Limit Exceeded)

int main(void) {
    int n,count = 0,flag = 0;
    scanf("%d",&n);
    int a[n],b[n];
    for (int i = 0; i < n; i++) {
        scanf("%d%d",&a[i],&b[i]);
    }
    for (int i = 0; i < n; i++) {
        for (int j = a[i]; j <= b[i]; j++) {
            flag = 0;
            if (j % 2 == 0 || j == 2 || j == 1) {
                continue; 
            }
            for (int k = 2; k < j; k++) {
                if (j % k == 0) {
                    flag = 1;
                } 
            }
            for (int k = 2; k < (j + 1) / 2 ; k++) {
                if (((j + 1) / 2) % k == 0) {
                    flag = 1;
                }
            }
            if (!flag) {
                count++;
            }
        }
        printf("%d\n",count);
        count = 0;
    }
    return 0;
}