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 <stdlib.h>
//#include <string.h>
#include <stdbool.h>
//#include <limits.h>
#include <math.h>

bool primeCheck(int m)
{
  if (m == 1)
    return false;
  if (m == 2)
    return true;
  bool b = true;
  int i;
  for (i = 2; i < sqrt((double)m); i++)
  {
    if (m % i != 0)
      continue;
    b = false;
    break;
  }
  if (m % i == 0)
    b = false;
  return b;
}


int main() {
  int q;
  scanf("%d", &q);
  int l[q], r[q];
  for (int i=0; i<q; i++)
    scanf("%d%d", l+i, r+i);
  int f[50001];
  f[0] = 0;
  for (int i=1; i<50000; i++) {
    if (primeCheck(2*i-1)) {
      if (primeCheck(i)) {
        f[i] = f[i-1]+1;
        continue;
      }
    }
    f[i] = f[i-1];
  }
  for (int i=0; i<q; i++) {
    printf("%d\n", f[(r[i]+1)/2] - f[(l[i]-1)/2]);
  }
  return 0;
}