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, n) for(int i = 0; i < n; ++i)
#define REPP(i, n) for(int i = 1; i <= n; ++i)
#define REPR(i, n) for(int i = n-1; i >= 0; --i)
#define FOR(i, m, n) for(int i = m; i < n; ++i)
#define ALL(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
const int MOD = 1000000007;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b) {if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b) {if(a<b) a=b;}

template <typename T>
bool isPrime(T x) {
    if(x==1) return false;
    for(T i = 2; i*i <= x; ++i) {
        if(x%i==0) return false;
    }
    return true;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    vector<int> v(100010, 0), s(100010);
    s[0]=0;
    REPP(i,v.size()) {
        if(i%2==1 && isPrime(i) && isPrime((i+1)/2)) ++v[i];
        s[i] = s[i-1]+v[i];
    }
    int q; cin >> q;
    int ans;
    REP(i,q) {
        int l,r; cin >> l >> r;
        ans = s[r]-s[l-1];
        cout << ans << '\n';
    }
    return 0;
}