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>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;
#define rep(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define is(a, b) a == b
#define len(v) ll(v.size())

const ll mod=1e9+7;

//vector書き出し
template <class T>
void cout_vec(const vector<T> &vec1){
  rep(i,len(vec1)){
    cout<<vec1[i]<<' ';
  }
  cout<<'\n';
}

template <class T>
bool is_prime(T x){
    for(T i=2;i*i<=x;i++){
        if(x%i==0){
            return false;
        }
    }
    return x!=1;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int q;
    cin>>q;
    vector<int> sum(100010,0);
    FOR(i,2,len(sum)){
        if(is_prime(i) && is_prime((i+1)/2)){
            sum[i]=sum[i-1]+1;
        }else{
            sum[i]=sum[i-1];
        }
    }
    rep(i,q){
        int l,r;
        cin>>l>>r;
        //cout<<sieve(r)<<' '<<sieve(l-1)<<'\n';
        cout<<sum[r]-sum[l-1]<<'\n';
    }
}