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: Runtime Error)

#include<stdio.h>

int prime_check[10000];

int isPrime(int a) {
  if (prime_check[a] == 1) {
    //printf("test1 %d\n", a);
    return 1;
  }

  if (prime_check[a] == -1) {
    return 0;
  }

  if (a==1) {
    prime_check[a] = -1;
    return 0;
  }

  if (a>2 && a%2 == 0) {
    prime_check[a] = -1;
    //printf("test2 %d\n", a);
    return 0;
  }

  for (int i=3; i<a/2; i+=2) {
    if (a%i == 0) {
      prime_check[a] = -1;
      //printf("test3 %d\n", a);
      return 0;
    }
  }

  prime_check[a] = 1;
  //printf("test4 %d\n", a);
  return 1;
}

int main(void) {
  int n;
  
  scanf("%d", &n);

  for (int i=0; i<n; i++) {
    int l, r;
    scanf("%d %d", &l, &r);

    int count = 0;
    for (int j = l; j <= r; j++) {
      int m = (j+1) / 2;
      //printf("==%d==\n", j);

      if (isPrime(j)==1 && isPrime(m)==1) {
        count++;
      }
    }

    printf("%d\n", count);
  }
}