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;
using Graph = vector<vector<int>>;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define Rep(i, k, n) for (int i = k; i < (int)(n); i++)
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; }
const long long INF = 1LL << 60;
long long dp[100010];
long long h[100010];
const int MOD = 1000000007;

int main()
{
   int MAX = 101010;
   vector<int> is_prime(MAX, 1);
   is_prime[0] = 0, is_prime[1] = 0;
   Rep(i, 2, MAX)
   {
       if (!is_prime[i]) continue;
       for (int j = i * 2; j < MAX; j += i) is_prime[j] = 0;
   }

   vector<int> a(MAX, 0);
   rep(i, MAX)
   {
       if (i % 2 == 0) continue;
       if (is_prime[i] && is_prime[(i + 1) / 2]) a[i] = 1;
   }

   vector<int> s(MAX + 1, 0);
   rep(i, MAX) s[i + 1] = s[i] + a[i];

   int q; cin >> q;
   rep(i, q)
   {
       int l, r;
       cin >> l >> r;
       r++;

       cout << s[r] - s[l] << endl;
   }
}