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>
#include <string.h>

int main(){
  int q,i,j,a;
  int l[100004],r[100004];
  int ll,rr;
  scanf("%d",&q);
  for(i=0;i<q;i++){
    scanf("%d %d",&l[i],&r[i]);
  }
  int x[300004];
  x[0]=1;
  x[1]=1;
  for(i=2;i<=300002;i++){
    if(x[i]==0){
      for(j=2*i;i<=300002-j;j+=i){
        x[j]=1;
      }
    }
  }
  int y[100004];
  y[0]=0;
  for(i=1;i<100004;i++){
    if((x[2*i-1]==0)&&(x[i]==0)){
      y[i]=1;
    }
  }
  for(i=1;i<100004;i++){
    y[i]+=y[i-1];
  }
  for(i=0;i<q;i++){
    ll=(l[i]+1)/2-1;
    rr=(r[i]+1)/2;
    a=y[rr]-y[ll];
    printf("%d\n",a);
  }
  return 0;
}