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>
int univprime[100000] = {};
int univcnt = 0;
void primemake(){
  int i;
  int j=0;
  int primef = 1;
  for(i=2;i<100000;i++){
    primef = 1;
    for(j=0;j<univcnt;j++){
      if(i % univprime[j] == 0){
        primef = 0;
        break;
      }
    }
    if(primef){
      univprime[univcnt] = i;
      univcnt++;
    }
  }
}

int main(){
  int hoge[50000] = {};
  int q = 0;
  int l[100000] = {};
  int r[100000] = {};
  int i = 0;
  int j = 0;
  int s = 0;
  int scheck = 0;
  int t = 0;
  int tcheck = 0;
  int likeflag = 0;
  int lp = 0;
  int rp = 0;
  int result = 0;
  scanf("%d",&q);
  for(i=0;i<q;i++){
    scanf("%d %d",&l[i],&r[i]);
  }
  
  primemake();
  
  for(i=0;i<50000;i++){
    s = 2*i + 1;
    t = (s + 1) / 2;
    scheck = 0;
    tcheck = 0;
    for(j=0;j<univcnt;j++){
      if(s == univprime[j]){
        scheck = 1;
      }
      if(t == univprime[j]){
        tcheck = 1;
      }
    }
    
    if(scheck*tcheck){
      //printf("%d/",i);
      for(j=i;j<50000;j++){
        hoge[j]++;
      }
    }
    
  }
  
  for(j=0;j<50000;j++){
    //printf("%d,",hoge[j]);
  }
  
  for(i=0;i<q;i++){
    lp = (l[i]-1)/2;
    rp = (r[i]-1)/2;

    result = hoge[rp] - hoge[lp];
    if(lp != 0 && hoge[lp-1] < hoge[lp]){
      result++;
    }
    printf("%d\n",result);
  }
  
  return 0;
}