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 <iostream>
#include <fstream>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>
#include <map>
#include <iomanip>
#include <stdlib.h>
#include <stdio.h>
#include <queue>
#include <deque>
#include <set>
#include <stack>
#include <time.h>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int, int> Pii;
typedef pair<int, ll> Pil;
typedef pair<ll, ll> Pll;
typedef pair<ll, int> Pli;

#define fi first
#define se second
#define mp make_pair
 
const ll MOD = 1e9 + 7;
const ll MOD2 = 998244353;
const ll INF = 1ll << 60;
const double PI = 2 * asin(1);

void yes() {printf("yes\n");}
void no() {printf("no\n");}
void Yes() {printf("Yes\n");}
void No() {printf("No\n");}
void YES() {printf("YES\n");}
void NO() {printf("NO\n");}

bool Prime[int(1e5 + 5)];
int Sum[int(1e5 + 5)], Q, L, R;

int main(){
  Prime[2] = true;
  for (int i = 3; i < 1e5 + 5; i++){
    Sum[i] = Sum[i - 1];
    if (i % 2 == 0) continue;

    Prime[i] = true;
    for (int j = 2; j * j <= i; j++){
      if (i % j == 0) Prime[i] = false;
    }
    
    if (Prime[i] && Prime[(i + 1) / 2]) Sum[i]++;
  }

  scanf("%d", &Q);
  for (int i = 0; i < Q; i++){
    scanf("%d%d", &L, &R);
    cout << Sum[R] - Sum[L - 1] << endl;
  }
  

  return 0;
}