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: Wrong Answer)

/*    特徴    */

/*    教訓    */

#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using ld  = long double;
using pint = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
const int INF=1e9+7;
const ll LINF=9223372036854775807;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld EPS = 1e-9; //微調整用（EPSより小さいと0と判定など）

int ii() { int x; if (scanf("%d", &x)==1) return x; else return 0; }
long long il() { long long x; if (scanf("%lld", &x)==1) return x; else return 0; }
string is() { string x; cin >> x; return x; }
char ic() { char x; cin >> x; return x; }
void oi(int x) { printf("%d ", x); }
void ol(long long x) { printf("%lld ", x); }
void od_nosp(double x) { printf("%.15f", x); } // 古い問題用
void od(double x) { printf("%.15f ", x); }
// long doubleで受け取り、fをLfなどに変えて出力すると、変な数値が出る
// それをなんとかするには独自の出力を作らなければならなそう
void os(const string &s) { printf("%s ", s.c_str()); }
void oc(const char &c) { printf("%c ", c); }
#define o_map(v){cerr << #v << endl; for(const auto& xxx: v){cout << xxx.first << " " << xxx.second << "\n";}} //動作未確認
void br() { putchar('\n'); }

#define gcd __gcd //llは受け取らない C++17~のgcdと違うので注意
int lcm(int a, int b){return a / gcd(a, b) * b;}
#define begin_end(a) a.begin(),a.end() //sort(begin_end(vec));
#define REP(i,m,n) for(ll i=(ll)(m) ; i < (ll) (n) ; i++ )
#define rep(i,n) REP(i,0,n)
#define m_p(a,b) make_pair(a,b)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define p_b push_back
#define SZ(x) ((int)(x).size) //size()がunsignedなのでエラー避けに

//4近傍（上下左右） rep(i, 2) にすると右・下だけに進む
vector<int> dx_4 = {1, 0, -1, 0};
vector<int> dy_4 = {0, 1, 0, -1};

// coutによるpairの出力（空白区切り）
template<typename T1, typename T2> ostream& operator<<(ostream& s, const pair<T1, T2>& p) {return s << "(" << p.first << " " << p.second << ")";}
// coutによるvectorの出力（空白区切り）
template<typename T> ostream& operator<<(ostream& s, const vector<T>& v) {
  int len = v.size();
  for (int i = 0; i < len; ++i) {
    s << v[i]; if (i < len - 1) s << " "; //"\t"に変えるとTabで見やすく区切る
  }
  return s;
}
// coutによる多次元vectorの出力（空白区切り）
template<typename T> ostream& operator<<(ostream& s, const vector< vector<T> >& vv) {
  int len = vv.size();
  for (int i = 0; i < len; ++i) {
    s << vv[i] << endl;
  }
  return s;
}

//最大値、最小値の更新。更新したor等しければtrueを返す
template<typename T>
bool chmax(T& a, T b){return (a = max(a, b)) == b;}
template<typename T>
bool chmin(T& a, T b){return (a = min(a, b)) == b;}

// -------- template end - //


// - library ------------- //

  // エラトステネスの篩
  // O(N log(logN)) で isPrime[0～n]のvectorを返す
  // 普通の素数判定を使うと O(N √N) なので、N=1000000 くらいならこっちが安全
  // 使い方 : vector<bool> isPrime = Eratosthenes(n);
  // isPrime(0), isPrime(1) はもちろん false を返す
  // 「素数の数」を返したければ、return p; とすればいい
  vector<bool> Eratosthenes(ll n){
    vector<ll> prime(n);
    vector<bool> isPrime(n + 1);
    ll p = 0; // n 以下の素数の数
    rep(i, n+1){
      isPrime.at(i) = true;
    }
    isPrime.at(0) = false;
    isPrime.at(1) = false;
    REP(i, 2, n+1){
      if (isPrime.at(i)){
        p++;
        prime.at(p) = i;
        for(ll j = 2*i; j <= n; j += i){
          isPrime.at(j) = false;
        }
      }
    }
    return isPrime;
  };

// --------- library end - //


int main(){


  vector<bool> isPrime = Eratosthenes(100001);



  vector<ll> cumsum(100001, 0);
  REP(i, 1, 100001){
    cumsum[i] = cumsum[i-1];
    if (i % 2 == 1 && isPrime[i] && isPrime[(i+1)/2]) cumsum[i]++;
  }

  ll Q = il();
  rep(i, Q){
    ll l = il();
    ll r = il();
    cout << cumsum[r] - cumsum[l-1] << endl;
  }
  
  cout << endl;
  cout << isPrime[0] << " " << isPrime[1] << endl;

}