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 Kotlin (1.0.0) (judge verdict: Wrong Answer)


fun main(args: Array<String>) {
    var r= readLine()
    var c= r!!.toInt()
    var array= mutableListOf<MutableList<Int>>()
    for(i in Array(c,{it})){
        var r= readLine()
        var array2=r!!.split(" ")
        var array3= mutableListOf<Int>()
        for(j in array2){
            array3.add(j.toInt())
        }
        array.add(array3)
    }
    for(i in array){
        var k=soCount(Array((i[1]-i[0])/2+1,{it}))
        println(k)
    }
}

fun so(a:Int):Boolean{
    if(a==1)return false
    for(i in Array(a-2,{it+2})){
        if(a%i==0)return false
    }
    return true
}

fun soCount(a:Array<Int>):Int{
    var c=Array(a.max()!!,{false})
    for(i in Array(a.max()!!,{it})){
        if(i==0||i==1||!a.contains(i)||!so((i+1)/2)){
            c[i]=false
            continue
        }
        for(j in Array(a.max()!!,{it})){
            for(k in Array(isqrt(a.max()!!)+1,{it})){
                if(j == i*k && c[j]){
                    c[j]=false
                }
            }
        }
    }
    return c.count({true})
}

fun isqrt(n:Int):Int{
    if(n<=0)return 0
    if(n<4)return 1
    var x=0
    var y=n
    while(x!=y&&x+1!=y){x=y;y=(n/y+y)/2}
    return x
}