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 <math.h>
#include <stdbool.h>
#include <stdio.h>
bool isLike2017(int n) {
	for (int j = 0; j < 2; j++) {
		for (int i = 2; i <= (int)sqrt(n); i++) {
			if (n % i == 0) {
				return false;
			}
		}
		n = (n + 1) / 2;
	}
	return true;
}
int main() {
	int q, l, r, list[100000], num, sum[100000];
	scanf("%d", &q);
	for (int i = 0; i < 100000; i++) {
		if (i % 2 == 0 || i < 3) {
			list[i] = 0;
			continue;
		}
		if (isLike2017(i)) {
			list[i] = i;
			num++;
		} else {
			list[i] = 0;
		}
	}
	sum[0] = 0;
	for (int i = 1; i < 100000; i++) {
		sum[i] = sum[i - 1];
		if (list[i] != 0) {
			sum[i]++;
		}
	}
	for (int i = 0; i < q; i++) {
		scanf("%d%d", &l, &r);
		printf("%d\n", sum[r] - sum[l - 1]);
	}
	return 0;
}