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,a,b) for(int i=a;i<(b);++i)
#define erep(i,a,b) for(int i=a;i<=(int)(b);++i)
#define per(i,a,b) for(int i=(a);i>(b);--i)
#define eper(i,a,b) for(int i=(a);i>=b;--i)
#define pb push_back
#define mp make_pair
#define INF 100100100100
#define MOD 1000000007
#define ALL(x) begin(x),end(x)
using namespace std;
typedef long long ll;
typedef pair<int,int> Pii;
typedef vector<int> vii;
typedef vector<ll> vll;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a, b)*b;}
bool isPrime(int x) {
  if (x < 2)
    return false;
  else if (x == 2)
    return true;
  else if (x % 2 == 0)
    return false;

  double sq = sqrt(x);
  for (int i = 3; i <= sq; i += 2) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}

int q;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
    cin >> q;
    vii p;
    rep(i, 0, 100001) {
      if (isPrime(i) && isPrime((i+1) / 2)) {
        p.pb(i);
      }
    }
    rep(i, 0, q) {
      int l, r;
      cin >> l >> r;
      cout << distance(lower_bound(ALL(p), l), upper_bound(ALL(p), r)) << endl;
    }
    return 0;
}