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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#include <numeric>
typedef long long Int;
#define rep(i,a,b) for(Int i=a;i<b;++i)
#define rrep(i,a,b) for(Int i=a;i>=b;--i)
using namespace std;

void SieveEratosthenes(long long n, vector<bool>& isPrime) {
    if (n+1 > isPrime.size()) isPrime.resize(n+1, true);
    isPrime[0] = isPrime[1] = false;

    for(long long i = 2; i*i <= n; ++i)
        if(isPrime[i])
            for(long long j = 2; i*j <= n; ++j)
                isPrime[i*j] = false;
}

int main() {
    cin.tie(0); cout.tie(0);
    ios::sync_with_stdio(false);

    Int Q; cin >> Q;
    const Int N = 100001;
    vector<bool> isPrime(N, true);
    vector<Int> subsum(N, 0);

    SieveEratosthenes(N, isPrime);

    rep(i, 1, N) {
        subsum[i] = subsum[i-1];
        if (i%2 == 1 and isPrime[i] and isPrime[(i+1)/2])
            subsum[i]++;
    }

    rep(i, 0, Q) {
        Int r, l; cin >> l >> r;
        cout << subsum[r] - subsum[l-1] << endl;
    }
}