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 Julia (1.4.0) (judge verdict: Wrong Answer)

parseInt(x) = parse(Int, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)


prime_memo = Dict{Int, Bool}()
like_memo = Dict{Int, Bool}()

function isPrime(n)
    if haskey(prime_memo, n)
        return prime_memo[n]
    end

    if n == 1
        return false
    end
    for i in 2:sqrt(n)
        if n % i == 0
            prime_memo[n] = false
            return false
        end
    end
    prime_memo[n] = true
    return true
end


function isLike(n)
    if haskey(like_memo, n) 
        return like_memo[n]
    end
    like_memo[n] =  isPrime(n) && isPrime((n + 1) ÷ 2)
    return like_memo[n] 
end



function main()
    A = Array{Int}(undef, 10^5)
    for i in 1:10^5 - 1
        if isLike(i)
            A[i + 1] = A[i] + 1
        else
            A[i + 1] = A[i]
        end
    end

    Q = parseInt(readline())
    for _ in 1:Q
        l, r = parseMap(split(readline()))
        if l == 1
            println(A[r])
        else
            println(A[r] - A[l - 1])
    
        end

    end
end


main()