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: Accepted)

#include<stdio.h>

int sosu(int a){
  int i = 2 , flag = 1;
  while(i*i <= a){
    if(a%i == 0){
      flag = 0;
      i += 10000000;
    }
    i ++;
  }

  return flag;
}

int main(void){
  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 k[100010] = {0};
  for(int i = 3;i <= 100000;i += 2){
    int a = sosu(i);
    int b = sosu((i+1)/2);
    if(a == 1 && b == 1){
      //printf("%d\n",i);
      k[i] ++;
    }
  }

  for(int i = 1;i <= 100000;i ++){
    k[i+1] += k[i];
  }

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

  return 0;
}