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<stdlib.h>
#include<limits.h>
#include<math.h>
#define N 1000000

short int Prime[N];
short int like2017[N];
//素数なた1が, 素数でないなら-1が入る

void setPrime(){
	int i,j;
	for(i=0;i<N;i++)Prime[i] = 1;
	Prime[0] = -1;
	Prime[1] = -1;
	Prime[2] = 1;
	for(i=2;i*2<N;i++){
		Prime[2*i] = -1;
	}
	for(i=3;i<sqrt(N)+1;i+=2){
		if(Prime[i] == 1){
			for(j=3;j<sqrt(N)+1;j+=2){
				Prime[i*j] = -1;
			}
		}
	}
}

void set2017(){
	int i;
	for(i=0;i<N;i++)like2017[i] = -1;
	for(i=3;i<sqrt(N)+1;i+=2){
		if(Prime[i] == 1 && Prime[(i+1)/2] ==1)like2017[i] = 1;
		/*
		if(Prime[i] == 1){
			if(Prime[(i+1)/2] == 1)like2017[i] = 1;
			else like2017[i] = -1;
		}else like2017[i] = -1;
		*/
	}
}

int main(){
	int i,j,k,Q,l,r;
	int cnt;
	setPrime();
	set2017();
	scanf("%d",&Q);
	
	for(i=0;i<Q;i++){
		cnt = 0;
		scanf("%d",&l);
		scanf("%d",&r);
		for(j=l;j<r+1;j+=2){
			if(like2017[j] == 1)cnt++;
			/*
			if(Prime[i] == 1)
				if(Prime[(i+1)/2] == 1)cnt++;
			*/
		}
		printf("%d\n",cnt);
	}
	
	return 0;
}