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: Accepted)

#include <bits/stdc++.h>
#define MAX 100010
using namespace std;

int mark[MAX], counter = 0;
int bit[MAX];
bool isPrime[MAX];

void genSieve()
{
    memset(isPrime, false, sizeof(isPrime));

    int limit = sqrt(MAX) + 1;

    mark[1] = 1;
    for(int i = 4; i < MAX; i += 2){
        mark[i] = 1;
    }

    isPrime[2] = true;

    for(int i  = 3; i < MAX; i += 2){
        if(!mark[i]){
            isPrime[i] = true;
            if(i <= limit){
                for(int j = i * i; j < MAX; j += i * 2){
                    mark[j] = 1;
                }
            }
        }
    }
}


void update(int idx, int value)
{
    while(idx <= MAX){
        bit[idx] += value;
        idx += (idx & -idx);
    }
}

int query(int idx)
{
    int sum = 0;
    while(idx > 0){
        sum += bit[idx];
        idx -= (idx & -idx);
    }

    return sum;
}

int main()
{
    genSieve();
    int q, l, r;

    for(int i = 1; i < MAX; i++){
        if(isPrime[i] && isPrime[(i + 1) / 2]){
            update(i, 1);
        }
    }

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

    return 0;
}