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<math.h>
int prime(int n)
{
    if (n < 2)
        return 0;
    if (n == 2)
        return 1;
    if (n % 2 == 0)
        return 0;
    int i;
    for (i = 3; i <= sqrt(n); i += 2)
        if (n%i == 0)
            return 0;
    return 1;
}
int main()
{
    int q;
    scanf("%d", &q);
    int i;
    int l[100005], r[100005];
        scanf("%d %d", &l[i], &r[i]);
    int p[100005];
    for (i = 0; i < 100005; i++)
        p[i] = 0;
    for (i = 0; i < 100005; i++)
        if (i % 2 == 1)
            if (prime(i) == 1 && prime((i + 1) / 2) == 1)
                p[i]++;
    for (i = 1; i < 100005; i++)
        p[i] += p[i - 1];
    for (i = 0; i < q; i++)
        printf("%d\n", p[r[i]] - p[l[i] - 1]);
    return 0;
}