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<bits/stdc++.h>
using namespace std;
int main(){
  int rp[100001] = {},n,i,j,l,r;cin >> n;bool su[100001] = {};
  vector<int> x;
  for(i=2;i<=100000;i++){
    rp[i] = rp[i-1];l = 1;
    for(auto j:x){
      if(j*j>i) break;
      if(i%j==0){
        su[i]++;break;
      }
    }
    su[i] ^= 1;
    if(su[i]) x.push_back(i);
    if(su[i] && i%2 && su[(i+1)/2]) rp[i]++;
  }
  while(cin >> l >> r) cout << rp[r]-rp[l-1] << "\n";
  return 0;
}