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: Wrong Answer)

#include<stdio.h>
#include<math.h>

int main(){
  int p[1000001],c[100001]={},i,j,q,r,l;
  for(i=2;i<=100000;i++){
    p[i]=1;
  }
  for(i=2;i<=317;i++){
    for(j=i;j<=317;j++){
      p[i*j]=0;
    }
  }
  for(i=3;i<100001;i+=2){
    c[i]+=p[i]*p[(i+1)/2];
  }
  for(i=3;i<100001;i++){
    c[i]+=c[i-1];
  }
  scanf("%d",&q);
  for(i=0;i<q;i++){
    scanf("%d%d",&l,&r);
    printf("%d\n",c[r]-c[l-1]);
  }
}