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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#include<math.h>
#define FOR(n) for(int i = 0;i < n;i++)
#define PRN(n) printf("%d\n",n)
int sosu(int a){
  if(a < 2)return 0;
  for(int j = 2;j <= (int)pow(a,0.5);j++){
    if(a % j < 1)return 0;
  }
  return 1;
}
int main(void){
  int n,l,r,f[100001],cnt = 0,x[100001];
  scanf("%d",&n);
  FOR(100001){
    f[i] = sosu(i);
  }
  FOR(100001){
    if(f[i] && i % 2 > 0){
      if(f[(i + 1) / 2]){
        cnt++;
      }
    }
    x[i] = cnt;
  }
  FOR(n){
    scanf("%d %d",&l,&r);
    PRN(x[r] - x[l - 1]);
  }
  return 0;
}