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)

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

#define NUM 100000

int prime[NUM + 1];

void init_prime() {
  int i, j;
  int sq_num = (int)sqrt((double)NUM);

  for (i = 0; i < NUM; i++) {
    prime[i] = i % 2;
  }
  prime[0] = prime[1] = 0;
  prime[2] = 1;

  for (i = 3; i < sq_num; i++) {
    if (prime[i] == 0) {
      continue;
    }
    for (j = i; i * j <= NUM; j += 2) {
      prime[i * j] = 0;
    }
  }
}

void solv(int q) {
  int i, j, c, l, r;

  for (i = 0; i < q; i++) {
    scanf("%d %d", &l, &r);
    c = 0;
    for (j = l; j <= r; j += 2) {
      if (prime[j] == 1 && prime[(j + 1) / 2] == 1) {
        c++;
      }
    }
    printf("%d\n", c);
  }
}

int main(int argc, char **argv) {
  int q;

  init_prime();
  while (scanf("%d", &q) != EOF) {
    solv(q);
  }

  return 0;
}