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: Wrong Answer)

#include <iostream>

#include <vector>
#include <map>
#include <algorithm>
#include <fstream>
#include<cstdio>
#include<iomanip>
#include<stack>
#include<queue>
#include<string>
#include <cstdlib>
#include <typeinfo>
#include <math.h>
#include <list>

#define REP(i, n) for(int i=0;i<n;i++)
#define REP2(i, s, n) for(int i=s;i<n;i++)
#define REP_1(i, n) for(int i=1;i<n+1;i++)
#define bitSearch(bit, n) for(int bit = 0; bit < (1 << n); bit++)
#define sz(x) int(x.size())
using namespace std;

template<class T>
void print(const T &value) {
    std::cout << value << std::endl;
}

void yesno(bool a) { if (a)cout << "yes" << endl; else cout << "no" << endl; }

void YesNo(bool a) { if (a)cout << "Yes" << endl; else cout << "No" << endl; }

void YESNO(bool a) { if (a)cout << "YES" << endl; else cout << "NO" << endl; }

typedef long long ll;
typedef unsigned long ul;
typedef long double ld;

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

ll INF = 1000001000;
ll mod = 1000000007;//10^9+7
ll power(ll x, ll n, ll mod) {//x^nをmodで割った余り
    ll res = 1;
    if (n > 0) {
        res = power(x, n / 2, mod);
        if (n % 2 == 0) res = (res * res) % mod;
        else res = (((res * res) % mod) * x) % mod;
    }
    return res;
}

int dx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int dy[8] = {0, -1, 1, 0, -1, 1, -1, 1};

struct Edge {
    int to, cost;

    Edge(int to, int cost) : to(to), cost(cost) {};
};

using Graph = vector<vector<int>>;//隣接リスト:G[i]にはiと隣接する頂点が入るよ。
using GraphW = vector<vector<Edge>>;//重み付きの隣接リスト
using P = pair<int, int>;//BFSで利用。queueに入れる。
using lP = pair<ll, ll>;
using PP = pair<int, P>; //dijkstraで利用,priority_queueに入れる。
using p_queue = priority_queue<int, vector<int>, greater<int>>;//小さい順

class UnionFindTree {

public:
    UnionFindTree(int size) : memberSize(size) {
        par.resize(size * sizeof(ll));
        rnk.resize(size * sizeof(ll));
        diff_weight.resize(size * sizeof(ll));
        REP(i, size) {
            par[i] = i;
            rnk[i] = 0;
            diff_weight[i] = 0;
        }
    }

    int memberSize;

    vector<int> par;
    vector<int> rnk;
    vector<ll> diff_weight;

//n個のnodeで初期化する関数
    void init(ll n) {
        REP(i, n) {
            par[i] = i;
            rnk[i] = 1;
        }
    }

//木の根を求める。この際，親は根に更新される。
    int find(ll x) {
        if (par[x] == x) { return x; }
        else {
            int r = find(par[x]);
            diff_weight[x] += diff_weight[par[x]];
            return par[x] = r;
        }
    }

//xとyが同じ集合に属するか判定
    bool same(ll x, ll y) {
        return find(x) == find(y);
    }

//xとyの属する集合を融合
    void unite(ll x, ll y) {
        x = find(x);
        y = find(y);
        if (x == y) { return; }

        if (rnk[x] < rnk[y]) {
            par[x] = y;
        } else {
            par[y] = x;
            if (rnk[x] == rnk[y]) { rnk[x]++; }
        }
    }

    //xの重み
    ll weight(ll x) {
        find(x);
        return diff_weight[x];
    }

    ll diff(ll x, ll y) {
        return weight(y) - weight(x);
    }

    //重みごと更新。weight(y)-weight(x)=wとなるようにする
    void merge(int x, int y, int w) {
        w += weight(x);
        w -= weight(y);
        x = find(x);
        y = find(y);
        if (x == y) { return; }
        if (rnk[x] < rnk[y]) {
            par[x] = y;
            diff_weight[x] = -w;
        } else {
            par[y] = x;
            diff_weight[y] = w;
            if (rnk[x] == rnk[y]) { rnk[x]++; }
        }
    }
};

priority_queue<P, vector<P>, greater<P>> que;

ll gcd(ll a, ll b) {
    if (a < b) {
        ll tmp = a;
        a = b;
        b = tmp;
    }
    ll r = a % b;
    while (r != 0) {
        a = b;
        b = r;
        r = a % b;
    }
    return b;
}


//番号ズレ注意！！
int main() {
    int N;
    cin >> N;
    int A[N], B[N], C[N];
    REP(i, N) { cin >> A[i]; }
    REP(i, N) { cin >> B[i]; }
    REP(i, N) { cin >> C[i]; }
    sort(A, A + N);
    sort(B, B + N);
    sort(C, C + N);
    int BC[N + 1];//BC[i]:=B[j]がB[i]以上の，B[i]とC[j]のペア数
    fill(BC, BC + N + 1, 0);
    int j = N - 1;
    for (int i = N - 1; i >= 0; i--) {
        while (C[j] > B[i]) {
            j--;
            if (j == -1)break;
        }
        BC[i] += N - 1 - j + BC[i + 1];
    }
//REP(i,N){
//    cout << BC[i] << endl;}

    ll ans = 0;
    int l = 0;
    REP(i, N) {
        while (B[l] <= A[i]) {
            l++;
            if (l == N)break;
        }
        ans += BC[l];
    }
    print(ans);
}