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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

const int mod = 1000000007;
const int inf = 1000000000;
const int size = 100000;

vector<bool> prime(size + 1, true);
vector<int> p(size + 1, 0);

int main() {
    prime[0] = prime[1] = false;
    for (int i = 2; i * i <= size; i++) {
        if (prime[i]) {
            for (int j = i * i; j <= size; j += i) {
                prime[j] = false;
            }
        }
    }
    for (int i = 1; i <= size; i++) {
        p[i] = p[i - 1] + (prime[i] and prime[i / 2 + 1]);
    }

    int q;
    scanf("%d", &q);
    int l, r;
    for (int i = 0; i < q; i++) {
        scanf("%d %d", &l, &r);
        printf("%d\n", p[r] - p[l - 1]);
    }

    return 0;
}