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)

#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<map>
#include<string.h>
#include<complex>
#include<math.h>
#include<queue>
#include <functional>
#include<time.h>
#include <stack>
#include<iomanip>
using namespace std;
#define rep(i,a,n) for(int i=(a);i<(n);i++)
#define ll long long
#define llint long long int
#define reverse(v)  reverse(v.begin(), v.end());
#define Yes(ans) if(ans)cout<<"Yes"<<endl; else cout<<"No"<<endl;
#define YES(ans) if(ans)cout<<"YES"<<endl; else cout<<"NO"<<endl;
#define hei(a) vector<a>
#define whei(a) vector<vector<a>>
#define UF  UnionFind 
#define Pint pair<int,int>
#define keta(a)  fixed << setprecision(a)
constexpr auto INF = 100000000000;
constexpr auto mod = 1000000007;
//辞書順はnext_permutation( begin( v ), end( v ) );やで！
// lower_boundは二分探索
struct edge { int to, cost; };
long long modpow(long long a, long long n, long long mod) {
	long long res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

// mod. m での a の逆元 a^{-1} を計算するよ！
long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}


//aCbをmod.mで割った余りを求める
ll int c(ll int a, ll int b, ll int m) {
	ll int ans = 1;
	for (ll int i = 0; i < b; i++) {
		ans *= a - i;
		ans %= m;
	}
	for (ll int i = 1; i <= b; i++) {
		ans *= modinv(i, m);
		ans %= m;
	}
	return ans;
}

//sは初めの点、vは頂点size、dは最短距離を求める(初期化はINF),Gは有向グラフ
void dijkdtra(int s, int v, vector<int>& d, whei(edge)& G) {
	priority_queue<Pint, vector<Pint>, greater<Pint>> que;//firstが距離、secondが頂点
	d[s] = 0;//スタートの距離はもちろんzero
	que.push(Pint(0, s));
	while (!que.empty()) {
		Pint p = que.top();
		que.pop();
		int V = p.second;
		if (d[V] < p.first)continue;
		for (int i = 0; i < G[V].size(); i++) {
			edge e = G[V][i];
			if (d[e.to] > d[V] + e.cost) {
				d[e.to] = d[V] + e.cost;
				que.push(Pint(d[e.to], e.to));
			}
		}
	}
}

llint binary_search(vector<int>&s,llint a) {
	llint l = -1;
	llint r = (int)s.size();
	while (r - l > 1) {
		llint mid = l + (r - l) / 2;

		if (s[mid]>=a) r = mid;
		else l = mid;
	}
	return r;
}


int main() {
	llint n;
	cin >> n;
	hei(int) a(n), b(n), c(n);
	rep(i, 0, n)cin >> a[i];
	rep(i, 0, n)cin >> b[i];
	rep(i, 0, n)cin >> c[i];
	sort(a.begin(), a.end());
	sort(b.begin(), b.end());
	sort(c.begin(), c.end());
	llint ans = 0;
	int r = 0, l = 0;
	rep(i, 0, n) {
		while (l < n && a[l] < b[i])l++;
		while (r <n && c[r] <= b[i])r++;
		ans += l * (n-r);
	}
	cout << ans << endl;
	return 0;
}