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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#include<cctype>
using namespace std;
#define rep(i,n) for (int i=0;i<(n);i++)
#define all(v) (v).begin(),(v).end()
typedef long long int ll;
#define pi 3.1415926535897932384
#define E9 1000000000
#define eps 1e-4
#define pii pair<int,int>

vector< bool > prime_table(int n) {
  vector< bool > prime(n + 1, true);
  if(n >= 0) prime[0] = false;
  if(n >= 1) prime[1] = false;
  for(int i = 2; i * i <= n; i++) {
    if(!prime[i]) continue;
    for(int j = i + i; j <= n; j += i) {
      prime[j] = false;
    }
  }
  return prime;
}


int main(){
  int Q; cin >> Q;
  auto table = prime_table(100005);
  vector<int> tab(100005,0);
  int cnt = 0;
  rep(i,100005){
    if (table[i] && table[(i+1)/2]) cnt++;
    tab[i] = cnt;
  }

  rep(q,Q){
    int l, r;
    cin >> l >> r;
    cout << tab[r+1]-tab[l-1] << endl;
  }


  // cout << fixed << setprecision(10);
  
  return 0;
}