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;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
typedef vector<ll> Vector;
typedef vector<Vector> DVector;

#define fi          first
#define se          second
#define pb          push_back
#define INF         INT_MAX
#define bcnt        __builtin_popcount
#define all(x)      (x).begin(),(x).end()
#define dbg(x)      cout<<#x"="<<x<<endl
#define uni(x)      x.erase(unique(all(x)),x.end())
#define ub(x,y)     upper_bound(all(x),y)-x.begin()
#define lb(x,y)     lower_bound(all(x),y)-x.begin()
#define rep(i,n)    repl(i,0,n-1)
#define usort(x)    sort(all(x))
#define dsort(x)    sort(all(x),greater<ll>())
#define mkp(x,y)    make_pair(x,y)
#define mmax(x,y)   (x>y?x:y)
#define mmin(x,y)   (x<y?x:y)
#define maxch(x,y)  x=mmax(x,y)
#define minch(x,y)  x=mmin(x,y)
#define exist(x,y)  (find(all(x),y)!=x.end())
#define each(itr,v) for(auto itr:v)
#define repl(i,a,b) for(ll i=(ll)(a);i<=(ll)(b);i++)
// エラトステネスの篩 [1,n]
bool is_prime[INF];
void sieve(ll n){
  for(ll i=0;i<=n;i++)is_prime[i]=true;
  is_prime[0]=is_prime[1]=false;

  for(ll i=2;i<=n;i++){
    if(is_prime[i]){
      for(ll j=2*i;j<=n;j+=i)is_prime[j]=false;
    }
  }
}
int main(){
  cin.sync_with_stdio(false);
  sieve(1000000);
  Vector prs(100000+1,0);
  repl(i,1,100000)if(i % 2 == 1 && is_prime[i] && is_prime[(i+1) / 2])prs[i]++;
  repl(i,2,100000)prs[i] += prs[i-1];
  ll q;
  cin >> q;
  Vector ans(q);
  rep(i,q){
    ll l,r;
    cin >> l >> r;
    cout << prs[r] - prs[l-1] << endl;
  }
}