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.

Examples

Input

1
3 7


Output

2


Input

4
13 13
7 11
7 11
2017 2017


Output

1
0
0
1


Input

6
1 53
13 91
37 55
19 51
73 91
13 49


Output

4
4
1
1
1
2
import java.util.Arrays;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class Main {
    static final int LIMIT = 100_000;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int Q = sc.nextInt();
        int[] l = new int[Q];
        int[] r = new int[Q];
        for (int i = 0; i < Q; ++i) {
            l[i] = sc.nextInt();
            r[i] = sc.nextInt();
        }

        System.out.println(solve(l, r));

        sc.close();
    }

    static String solve(int[] l, int[] r) {
        boolean[] primes = buildPrimes();

        boolean[] similars = new boolean[LIMIT + 1];
        for (int i = 0; i < similars.length; ++i) {
            similars[i] = i % 2 != 0 && primes[i] && primes[(i + 1) / 2];
        }

        int[] prefixSums = new int[LIMIT + 1];
        int prefixSum = 0;
        for (int i = 0; i < prefixSums.length; ++i) {
            prefixSum += similars[i] ? 1 : 0;
            prefixSums[i] = prefixSum;
        }

        return IntStream.range(0, l.length).mapToObj(i -> String.valueOf(prefixSums[r[i]] - prefixSums[l[i] - 1]))
                .collect(Collectors.joining("\n"));
    }

    static boolean[] buildPrimes() {
        boolean[] primes = new boolean[LIMIT + 1];
        Arrays.fill(primes, true);
        primes[0] = false;
        primes[1] = false;

        for (int i = 2; i < primes.length; ++i) {
            if (primes[i]) {
                for (long j = (long) i * i; j < primes.length; j += i) {
                    primes[(int) j] = false;
                }
            }
        }

        return primes;
    }
}