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;
const int N=100001;
int main(){
int lps[N]={0,0}; bool notP[N]={true,true}, frag[N]={true,true};
for(int i=2; i*i<N; i++){
if(!frag[i]){
  notP[i]=false; frag[i]=true;
  for(int m=2*i; m<N; m+=i){ notP[m]=frag[m]=true; }
}  
}
for(int i=2; i<N; i++){ lps[i]=lps[i-1]; if(!notP[i]&&!notP[(i+1)/2]){ lps[i]++; } }
int Q, l, r; cin >> Q;
for(int i=0; i<Q; i++){ cin >> l >> r; cout << lps[r]-lps[l-1] << "\n"; } 
}