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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define Q_MAX 100000

int main(void) {
  int Q, *l, *r;
  int n[Q_MAX+1] = {}, p[Q_MAX+1] = {};
  scanf("%d", &Q);
  l = (int*)calloc(Q, sizeof(int));
  r = (int*)calloc(Q, sizeof(int));
  for (int i = 0; i < Q; i++) {
    scanf("%d %d", &l[i], &r[i]);
  }
  for (int i = 2; i <= Q_MAX; i++) {
    p[i] = 1;
  }

  for (int i = 2; i <= Q_MAX; i++) {
    if (p[i] && i <= Q_MAX / 2) {
      for (register int k = 2; i*k <= Q_MAX; k++) {
        p[i*k] = 0;
      }
    }
    n[i] = n[i-1] + (p[i] && p[(i+1)/2]);
  }

  for (int i = 0; i < Q; i++) {
    printf("%d\n", n[r[i]] - n[l[i]-1]);
  }

  return 0;
}