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;

class Main {
    private static final int MAXN = 100000;
    private static boolean[] isPrime = new boolean[MAXN + 1];

    private static void generatePrimes() {
        Arrays.fill(isPrime, true);
        isPrime[0] = isPrime[1] = false;
        for(int i = 2; i <= MAXN; ++i) {
            if(isPrime[i]) {
                for(int j = 2 * i; j <= MAXN; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }

    private static Scanner scanner = new Scanner(System.in);
    public static void main(String[] args) {
        generatePrimes();
        int[] cumulativeCount = new int[MAXN + 1];
        for(int i = 1; i <= MAXN; ++i) {
            cumulativeCount[i] = cumulativeCount[i-1];
            if(isPrime[i] && isPrime[(i + 1) / 2]) {
                ++cumulativeCount[i];
            }
        }
        int q = scanner.nextInt();
        StringBuilder ans = new StringBuilder();
        while(q-- > 0) {
            int l = scanner.nextInt();
            int r = scanner.nextInt();
            int cnt = cumulativeCount[r] - cumulativeCount[l - 1];
            ans.append(cnt + "\n");
        }
        scanner.close();
        System.out.println(ans.toString());
    }
}