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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int acs(const void *a, const void *b) { 
  return *(int*)a - *(int*)b;
} /* 1,2,3,4.. */
int des(const void *a, const void *b) {
  return *(int*)b - *(int*)a;
} /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b) { 
  return *(char*)a - *(char*)b;
} /* a,b,c,d.. */
int cmp_str(const void *a, const void *b) {
  return strcmp(*(const char **)a, *(const char **)b);
} /* aaa,aab.. */
#define min(a,b) (a < b ? a : b)
#define max(a,b) (a > b ? a : b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 100005
#define MOD 1000000007

typedef long long int lli;

int main(void) {
  int Q;
  scanf("%d", &Q);

  int l[MAX], r[MAX];
  rep(i, 0, Q) scanf("%d %d", &l[i], &r[i]);

  int p[MAX] = {0};
  int n[MAX] = {0};
  for ( int i = 2; i <= 100000; i++ ) {
    int ok = 1;
    for ( int j = 2; j < i; j++ ) {
      if ( i % j != 0 ) continue;
      ok = 0; break;
    }
    if ( ok ) p[i] = 1;
    if ( ok && p[(i+1)/2] ) n[i] = 1; 
  }

  
  int sum[MAX];
  sum[1] = 0;
  for ( int i = 1; i <= 100000; i++ ) {
    sum[i + 1] = sum[i] + n[i];
  }

  rep(i, 0, Q) {
    printf("%d\n", sum[r[i]+1] - sum[l[i]]);
  }
  return 0;
}