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>
#include<string.h>
#include<stdlib.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, k, max;
  struct num_t* num;
  int* cache;
  
  scanf("%d", &q);
  num = (struct num_t*)malloc(sizeof(struct num_t) * q);
  max = 1;
  for (i = 0; i < q; ++ i) {
    scanf("%d %d", &(num[i].l), &(num[i].r));
    if (num[i].r > max) {
      max = num[i].r;
    }
  }
  cache = (int*)malloc(sizeof(int) * (max + 1));
 
  for (i = 1; i <= max; ++ i) {
    cache[i] = -1;
  }
  
  for (i = 0; i < q; ++ i) {
    l = num[i].l;
    r = num[i].r;
    c = 0;
	for (j = l; j <= r; j += 2) {
      if (cache[j] == -1) {
        cache[j] = is_sosu(j);
      }
      if (cache[j]) {
        k = (j + 1) >> 1;
        if (cache[k] == -1) {
          cache[k] = is_sosu(k);
        }
        if (cache[k]) {
          ++ c;
        } 
      }
    }
    printf("%d\n", c);
  }
  fflush(stdout);
  free(num);
  free(cache);
  return 0;
}