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: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>
#define NUM 100001  // NUM以下の素数を求める

int main(void){
    // Your code here!
int Q;
scanf("%d", &Q);
int L[Q];
int R[Q];
  unsigned prime[NUM];
for (int i = 0; i < Q; ++i) {
  scanf("%d" "%d", &L[i],&R[i]);

  unsigned i, j;
  unsigned sq_num = (int)sqrt((double)NUM);

  // 1が立っているものが素数
  // 0が立っているものが合成数（素数ではない）

  for (i = 0; i < NUM; i++)
    prime[i] = 1;  // 全ての数に1を立てる
  prime[0] = 0;    // 1は素数ではない

  for (i = 1; i < sq_num; i++) {
    if (prime[i] == 1)          // prime[i]が素数なら
      for (j = (i+1); (i+1) * j <= NUM; j++)
        prime[(i+1) * j - 1] = 0;  // 素数の倍数は素数ではない
  }

}

int nita[NUM]= {};
int i;
  for (i = 1; i < NUM; i++){
    if(prime[i-1]==1 && prime[(i)/2] ==1 && i != 2){
        nita[i] = nita[i-1] + 1;}
    else{
        nita[i] = nita[i-1];}
}
int a;
//  for (i = 0; i < 100; i++){
//    printf("%d\n",nita[i]);
//}
  for (i = 0; i < Q; i++){
    a = nita[R[i]] - nita[L[i]-1];
    printf("%d\n",a);
}
return 0;
}