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 FOR(I,X,Y) for(long long (I)=(X);(I)<(Y);(I)++)
#define REP(I,X,Y) for(long long (I)=(Y)-1;(I)>=(X);(I)--)
#define ALL(X) (X).begin(),(X).end()
#define pb push_back
#define debug(x) cerr<<#x<<':'<<x<<endl;
#define DEBUG(v) cerr<<#v<<':';for(auto x:v)cerr<<x<<' ';cerr<<endl;
#define INF 1000000007
#define LINF 1000000000000000007
#define EPS 1e-9
typedef long long ll;

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

//データ数 < 2^(log2n)
constexpr ll log2n = 20;
constexpr ll sizeofn = 1 << log2n+1;
ll SegmentTree[sizeofn];
ll STconvert(ll i,ll j){
  return (1<<log2n)/(j-i+1) + i/(j-i+1);
}

ll STmerge(ll i,ll j){
  return i+j;
}

ll STquery(ll i,ll j){
  if(i > j)swap(i,j);
  REP(rep,0,log2n+1){
    ll p = 1 << rep;
    if(i%p == 0 && i+p-1 < j){
      return STmerge(STquery(i+p,j),SegmentTree[STconvert(i,i+p-1)]);
    }
    else if(i%p == 0 && i+p-1 <= j){
      return SegmentTree[STconvert(i,j)];
    }
  }
}

void STchange(ll i,ll j){
  SegmentTree[STconvert(i,i)] = j;
  FOR(rep,1,log2n+1){
    ll p = 1 << rep;
    SegmentTree[STconvert(i/p*p,i/p*p+p-1)] = STmerge(SegmentTree[STconvert(i/p*p,i/p*p+p/2-1)],SegmentTree[STconvert(i/p*p+p/2,i/p*p+p-1)]);
  }
}

signed main(){
  vector<bool> v = primenumbers(100010);
  FOR(i,3,100010){
    if(v[i] && v[(i+1)/2]){
      STchange(i,1);
    }
  }
  int Q;
  cin >> Q;
  int l,r;
  FOR(i,0,Q){
    cin >> l >> r;
    cout << STquery(l,r) << endl;
  }
}