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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

var lines = [];
var readline = require('readline');

var rl = readline.createInterface({
    input: process.stdin,
    output: process.stdout,
});

rl.on('line', function (x) {
    lines.push(x);
});

rl.on('close', function () {
    var Q = Number(lines[0]); lines.shift();
    var lr = lines.map(i => i.split(" ").map(i => Number(i)));

    var x = lr.concat().sort((a, b) => b[1] - a[1])
    var max = x[0][1];

    var sum = Array(max + 1).fill(0);

    sum[0] = 0; sum[1] = 1; sum[2] = 2; sum[3] = 3;
    for (var i = 4; i <= max; i++) {
        for (var j = 2; j <= Math.sqrt(i); j++) {

            if (i % j === 0) { break; }
            if (((i + 1) / 2) % j === 0) { break; }
            if (j === Math.floor(Math.sqrt(i))) { sum[i]++; }
        }
        sum[i] += sum[i - 1];
    }
    
    for (var i = 0; i < Q; i++) {
        if (lr[i][1] === lr[i][0]) { console.log(1) }
        else { console.log(sum[lr[i][1] - 1] - sum[lr[i][0] - 1]) }
    }

});