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=1e5+5;
int A[N],n,cnt,q,mx,mn,x,y,z;
bool prime(int l){
    if(l<=1) return 0;
    for(int i=2;i<=l/i;++i)
        if(l%i==0) return 0;
    return 1;
}
int main(){
    for(int i=1;i<N;++i){
     A[i]=(prime(i)&&prime((i+1)/2));
     A[i]+=A[i-1];
    }
    //for(int i=0;i<55;++i)
      //  cout<<i<<":"<<A[i]<<endl;
    scanf("%d",&q);
    while(q--){
        scanf("%d %d",&x,&y);
        printf("%d\n",A[y]-A[x-1]);
    }
    return 0;
}
