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)

// AtCoder template
// 見た奴全員saba!
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i = 0; i < n; ++i)

// 素数判定
// 計算量O(√n)
template<class T>
bool isPrime(T n){
    if(n < 2) return false;
    for(long long i = 2; i*i <= n; ++i) if(n % i == 0) return false;
    return true;
}


// 素数列挙
// nまでの素数をEratosthenesの篩で列挙する
// 計算量O(nloglogn)
template<class T>
vector<T> prime_list(T n){
    vector<T> ret;
    vector<T> isPrime(n+1,1);
    isPrime[0] = isPrime[1] = 0;
    for(long long i = 2; i <= n; ++i){
        if(isPrime[i] == 0) continue;
        ret.push_back(i);
        for(long long j = i*i; j <= n; j += i) isPrime[j] = 0;
    }
    return ret;
    // return isPrime;  // 状況によってはこっちを返す
}

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

    int m = 100001,q; cin >> q;

    vector<int> p = prime_list(m);
    int n = p.size();
    vector<bool> like_list(m+1,false);
    rep(i,n){
        if(isPrime((p[i]+1)/2)) like_list[p[i]] = true;
    }
    like_list[2] = false;
    vector<int> pn_sum(m+1);
    pn_sum[0] = 0;
    rep(i,m){
        if(like_list[i+1]) pn_sum[i+1] = pn_sum[i]+1;
        else pn_sum[i+1] = pn_sum[i];
    }
    
    int l,r;
    rep(i,q){
        cin >> l >> r;
        cout << pn_sum[r] - pn_sum[l-1] << "\n";
    }
}