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 main(){
	int i,j=0,k,q;
	int l[100000],r[100000],a[100000];
	int prime[10000];
	int like2017[1000];
	prime[0]=2;
	like2017[0]=3;
	
	for(i=3;i<100000;i+=2){
		for(k=0;k<=j;k++){
			if(i%prime[k]==0){
				break;
			}
			if(k==j){
				j++;
				prime[j]=i;
			}
		}
	}
	prime[j+1]=-1;
	j=0;
	
	for(i=2;prime[i]!=-1;i++){
		for(k=0;k<=i;k++){
			if((prime[i]+1)/2==prime[k]){
				j++;
				like2017[j]=prime[i];
				break;
			}
		}
	}
	like2017[j+1]=100000;
	
	scanf("%d",&q);
	for(i=0;i<q;i++){
		scanf("%d%d",&l[i],&r[i]);
	}
	
	for(i=0;i<q;i++){
		j=0;
		k=0;
		while(1){
			if(l[i]<=like2017[j]){
				break;
			}
			j++;
		}
		while(1){
			if(r[i]<like2017[j]){
				break;
			}
			k++;
			j++;
		}
		a[i]=k;
	}
	
	for(i=0;i<q;i++){
		printf("%d\n",a[i]);
	}
	
	return 0;
}