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 <iostream>
#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define F first
#define S second
#define fo(i,a,b)  for(int i = a; i < b; i++)

using ll = long long;
using ld = long double;
using ul = unsigned long long;
using ii = pair<int, int>;
#define Mosta     ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ReadFile            freopen("input.in","r",stdin)
#define WriteFile           freopen("output.txt","w",stdout)
const ll mod7 = 1000000007,mod = 1000000000,OO = 0x3f3f3f3f;
const ld EPS = 1e-7;
double Pi=acos(-1);
const ld rad=Pi/180.0;
ll power(ll x, ll y, ll p)
{
    ll res = 1;
    while (y > 0)
    {
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}
ll nCr (ll n,ll m)
{
        return(m == 0)?1:n*nCr(n-1,m-1)/m;
}
/*****************************************************/

const int N = (int) 1e5 + 5;

bool prime[N];

int ans[N];

int main()
{
    memset(prime , 1 , sizeof(prime));
    prime[0]=prime[1]=0;
    for(int i = 2; i <= N; ++i){
        if(prime[i]){
            for(int j = i+i; j <= N; j+=i)
                prime[j] = 0;
        }
    }

    for(int i = 3; i <= 100000; i+=2){
        if(prime[i] && prime[(i+1)/2])
            ans[i]++;
    }

    for(int i = 3; i <= 100000; i++)
        ans[i] += ans[i-1];

    int q;
    cin >> q;

    while(q--){
        int l , r;
        cin >> l >> r;
        cout << ans[r] - ans[l-1] << "\n";
    }

    return 0;
}