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 pair<ll, ll> P;

int Q, l[100001], r[100001];

int main(){
    cin >> Q;
    for(int i=0;i<Q;i++)    cin >> l[i] >> r[i];

    vector<int> nums(100001);
    for(int i=1;i<100001;i++){
        nums[i] = i;
    }

    vector<int> ans;
    for(int i=2;i<=100000;i++){
        if(nums[i] == 0) continue;
        for(int j=2;i*j<=100000;j++){
            nums[i*j] = 0;
        }
        int x = (i + 1) / 2;
        if(nums[x] != 0 && i != 2){
            ans.push_back(i);
        }
    }

    for(int i=0;i<Q;i++){
        auto li = lower_bound(ans.begin(), ans.end(), l[i]);
        auto ri = upper_bound(ans.begin(), ans.end(), r[i]);
        int cnt = ri - li;
        cout << cnt << endl;
    }
}