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.

Examples

Input

1
3 7


Output

2


Input

4
13 13
7 11
7 11
2017 2017


Output

1
0
0
1


Input

6
1 53
13 91
37 55
19 51
73 91
13 49


Output

4
4
1
1
1
2
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
int Q, pn[100001], pn2017[100001];
int main() {
  for(int i = 2; i <= 1e5; ++i) pn[i] = 1;
  for(int i = 2; i <= sqrt(1e5); ++i) {
    if(!pn[i]) continue;
    for(int j = 2; i*j <= 1e5; ++j) pn[i*j] = 0;
  }
  for(int i = 2; i <= 1e5; ++i) {
    pn2017[i] = pn2017[i-1]+(pn[i]&&pn[(i+1)/2]);
  }
  scanf("%d", &Q);
  for(int i = 1, l, r; i <= Q; ++i) {
    scanf("%d %d", &l, &r);
    printf("%d\n", pn2017[r]-pn2017[l-1]);
  }
}