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;
#define rep(i,n) for(int i=0; i<(n) ;i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
using ll=long long;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define MOD  1000000007
const ll INF=1e18;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b){ a = b; return 1; } return 0; }

bool isprime(long long int val) {
 if (val <= 1)return false;
 for (long long int i = 2; i*i <= val; i++) {
   if (val%i == 0LL)return false;
 }
 return true;
}

bool f[100010];
int cnt[100010];

int main(int argc, char const *argv[]) {
  int Q;
  cin >> Q;

  for(int i=2; i<=100000; i++){
    f[i]=isprime(i);
  }

  for(int i=1; i<=100000; i++){
    if(f[i] && f[(i+1)/2] ) cnt[i]=1;
  }

  vector<int> s(100010,0);

  for(int i=0; i<=100000; i++){
    s[i+1]=s[i]+cnt[i];
  }

  while(Q--){
    int l,r;
    cin >> l >> r;
    int ans = s[r+1] - s[l];
    
    cout << ans << endl;
  }

  return 0;
}