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: Time Limit Exceeded)

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

struct num_t {
  int l;
  int r;
};

int is_sosu(int n) {
  int i;
  double rn;
  if (n < 2) return 0;
  else if (n == 2) return 1;
  else if (n % 2 == 0) return 0;
  rn = sqrt(n);
  for (i = 3; i <= rn; i += 2) {
    if (n % i == 0) return 0;
  }
  return 1;
}

int main() {
  int q, l, r, c, i, j;
  struct num_t num[100000];
  scanf("%d", &q);
  for (i = 0; i < q; ++ i) {
    scanf("%d %d", &(num[i].l), &(num[i].r));
  }
 
  for (i = 0; i < q; ++ i) {
    l = num[i].l;
    r = num[i].r;
    c = 0;
	for (j = l; j <= r; j += 2) {
      if (is_sosu(j) && is_sosu((j + 1) / 2)) {
        ++ c;
      }
    }
    printf("%d\n", c);
  	fflush(stdout);
  }
  return 0;
}