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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
// 総数のあと改行して改行区切りの数字の集合
fscanf(STDIN, '%d', $q);
for ($i  = 0; $i < $q; $i++) {
    $lr[] = array_map('intval', explode(' ', trim(fgets(STDIN))));
}

$is_prime = is_prime(10 ** 5);

$s = array_fill(0, 10 ** 5, 0);
for ($i = 1; $i <= 10 ** 5; $i++) {
    if ($is_prime[$i] && $is_prime[($i + 1) / 2]) {
        $s[$i] = $s[$i - 1] + 1;
    } else {
        $s[$i] = $s[$i - 1];
    }
}

for ($i = 0; $i < $q; $i++) {
    $l = $lr[$i][0];
    $r = $lr[$i][1];
    echo ($s[$r] - $s[$l - 1]).PHP_EOL;
}

/**
 * エラトステネスのふるい
 * $maxまでの整数が素数かどうかboolを格納した配列を返す
 */
function is_prime($max) {
    $arr = array_fill(0, $max + 1, true);
    $arr[0]  = $arr[1] = false;
    $rmax = floor(sqrt($max));
    for ($i = 2; $i <= $rmax; $i++) {
        for ($j = 2 * $i; $j <= $max; $j += $i) {
            $arr[$j] = false;
        }
    }
    return $arr;
}