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>
int main(void){
  int q,lr[2][100000],isPrime[100001],d2017[100001],ans[100001],i,j;
  scanf("%d",&q);
  for(i=0;i<q;i++)scanf("%d%d",&lr[0][i],&lr[1][i]);
  for(isPrime[0]=0,isPrime[1]=0,i=0;i<100001;i++)isPrime[i]=1;
  for(isPrime[0]=0,isPrime[1]=0,i=2;i<317;i++)if(isPrime[i])for(j=(i+i);j<=100000;j+=i)isPrime[j]=0;
  for(d2017[0]=0,d2017[1]=0,i=1;i<=10001;i+=2)if(isPrime[i]&&isPrime[(i+1)/2])d2017[i]=1;
  for(ans[0]=0,i=1;i<=100000;i++)ans[i]=ans[i-1]+d2017[i];
  for(i=0;i<q;i++)printf("%d\n",ans[lr[1][i]]-ans[lr[0][i]-1]);
  return 0;
}