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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <map>
#include <utility>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <set>
#include <unordered_map>
#include <unordered_set>

#define REP(i,a,n) for(int i=a;i<n;++i)
#define rep(i,n) REP(i,0,n)
#define REV(i,a,n) for(int i=n;i>=a;--i)
#define all(e) e.begin(),e.end()
#define rall(e) e.rbegin(),e.rend()
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define show(n) cerr<<#n<<" = "<<n<<endl
#define shows(n) for(auto z:n){cerr<<z<<", ";}cerr<<endl

//ここからAtCoder向けdefine
#define yes cout << "YES" << endl
#define no cout << "NO" << endl

using namespace std;

using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using pint = pair<int,int>;

const int INF = ~(1 << 31);

int maxQ = 100001;
vector<bool> IsPrime;

void sieve(size_t max){
    if(max+1 > IsPrime.size()){     // resizeで要素数が減らないように
        IsPrime.resize(max+1,true); // IsPrimeに必要な要素数を確保
    } 
    IsPrime[0] = false; // 0は素数ではない
    IsPrime[1] = false; // 1は素数ではない

    for(size_t i=2; i*i<=max; ++i) // 0からsqrt(max)まで調べる
        if(IsPrime[i]) // iが素数ならば
            for(size_t j=2; i*j<=max; ++j) // (max以下の)iの倍数は
                IsPrime[i*j] = false;      // 素数ではない
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    sieve(maxQ);
    int q;
    cin >> q;
    int l, r;
    int ans;
    int i;
    while(~scanf("%d %d",&l,&r)){
        ans = 0;
        for(i = l;i <= r;i += 2){
            if(IsPrime[i] && IsPrime[(i + 1) / 2]){ ans++; }
        }
        printf("%d\n", ans);
    }

    return 0;
}