Problem:
Score : 300 points

Problem Statement

The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower.

He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i.

To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar.

How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9(1\leq i\leq N)

1 \leq B_i \leq 10^9(1\leq i\leq N)

1 \leq C_i \leq 10^9(1\leq i\leq N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N
B_1 ... B_N
C_1 ... C_N

Output

Print the number of different altars that Ringo can build.

Sample Input 1

2
1 5
2 4
3 6

Sample Output 1

3

The following three altars can be built:

Upper: 1-st part, Middle: 1-st part, Lower: 1-st part

Upper: 1-st part, Middle: 1-st part, Lower: 2-nd part

Upper: 1-st part, Middle: 2-nd part, Lower: 2-nd part

Sample Input 2

3
1 1 1
2 2 2
3 3 3

Sample Output 2

27

Sample Input 3

6
3 14 159 2 6 53
58 9 79 323 84 6
2643 383 2 79 50 288

Sample Output 3

87

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

//ABC077-C(Snuke Festival)
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

//中部のパーツのサイズは上部のパーツのサイズより真に大きく、下部のパーツのサイズは中部のパーツのサイズより 真に大きい

int n;
vector <int> a,b,c;

int main(){
    
    cin>>n;
    for(int i=0; i<3; i++){
        for(int j=0; j<n; j++){
            int t;
            cin>>t;
            if(i==0)a.push_back(t);
            else if(i==1)b.push_back(t);
            else c.push_back(t);
        }
    }
    
    //a[x]<b[y]<c[z]を満たすx,y,zを全て求めてその個数を出力すればよい
    //b[y]を固定すると、(a[x]<b[y]を満たすxの個数)*(b[y]<c[z]を満たすzの個数)
    //これを全てのb[y]について行ったものの和を出力すればよい
    //愚直に配列を全て見るとO(N^2)で間に合わないのでa,cをソートしておき二分探索
    sort(a.begin(),a.end());
    sort(c.begin(),c.end());
    long long ans=0;
    for(int y=0; y<n; y++){
        //末尾のイテレータからupper_boundのイテレータを引くと、あるkeyより大きい要素の個数
        //lower_boundのイテレータから先頭のイテレータを引くと、あるkeyより小さい要素の個数
        ans+=(lower_bound(a.begin(),a.end(),b[y])-a.begin())*(c.end()-upper_bound(c.begin(),c.end(),b[y]));
    }
    cout<<ans<<endl;
    return 0;
}