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>
//#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
//using namespace boost::multiprecision;
typedef long long int ll;
typedef long double ld;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const long long INF = 1LL << 60;
bool pairCompare(const pair<double,ll>& firstElof, const pair<double,ll>& secondElof){
    return firstElof.first < secondElof.first;
}
bool pairCompareSecond(const pair<double,ll>& firstElof, const pair<double,ll>& secondElof){
    return firstElof.second < secondElof.second;
}
#define MAX_N 100100
#define MOD 998244353
// 四方向への移動ベクトル
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
 
ll gcd(ll a,ll b){
    if (a%b==0)return(b);
    else return(gcd(b,a%b));
}
bool x[MAX_N];
vector<ll> num;
void eratosthenes(ll n) {
    int p = 0;
    for (ll i=0; i<=n; ++i) x[i] = true;
    x[0] = x[1] = false;
    for(int i=2; i<=n; ++i) {
        if(x[i]){
            num.push_back(i);
            for(int j=2*i; j<=n; j+=i) x[j] = false;
        }
    }
    return;
}

int main(){
    ll q;
    cin >> q;
    eratosthenes(100000);
    ll rui[100100];
    fill(rui,rui+100010,0);
    for (ll i=1;i<100010;i++){
        if (i%2!=0 and x[i] and x[((i+1)/2)]){
            rui[i]=rui[i-1]+1;
            continue;
        }
        rui[i]=rui[i-1];
    }
    for (ll i=0;i<q;i++){
        ll l,r;
        cin >> l >> r;
        cout << rui[r]-rui[l-1] << endl;
    } 
    return 0;
}