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>
#define REP(i,n) for(int i = 0;i < (n);i++)
#define P pair<int,int>
#define pb push_back
#define mk make_pair
using namespace std;
const int INF = 1 << 20;

typedef long long ll;



vector <int> Ehurui;
void Etenesu(int y){
  bool f[y+1] = {0};
  for(int i = 2;i <= sqrt(y);i++ ){
    if(f[i] == false){
      Ehurui.push_back(i);
      f[i] = true;
      for(int j = i;j <= y;j = j+i){
	f[j] = true;
      }
    }
  }
  for(int i = 2;i <= y;i++){
    if(!f[i])
      Ehurui.push_back(i);
  }
}



int main(){
  int q;
  cin >> q;
  Etenesu(100000);
  bool f[100001] = {0};
  REP(i,Ehurui.size()){
    f[Ehurui[i]] = true;
  }
  bool ff[100001] = {0};
  REP(i,100001){
    if(f[i] == true && f[(i+1)/2] == true)
      ff[i] = true;
  }
  int w[100001] = {0};
  w[3] = 1;
  for(int i = 4;i <= 100000;i++){
    if(ff[i])
      w[i] = w[i-1]+1;
    else
      w[i] = w[i-1];
  }
  REP(i,q){
    int r,l;
    cin >> r >> l;
    
    
    cout <<w[l]-w[r-1]  << endl;
   
  }
  return 0;
}