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 <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <tuple>
#include <cassert>
#include <functional>
#include <set>
#include <deque>
using namespace std;
typedef long long ll;
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
const int INF = 1<<29;
#define LNF __LONG_LONG_MAX__;
const ll MOD = 1000000007;
const int MAX = 200000;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
// 法mでのaの逆元を計算
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;
}
// a^n mod を計算する
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;
}
//最大公約数
int gcd(int x, int y) { return (x % y)? gcd(y, x % y): y; }
//最小公倍数
int lcm(int x, int y) { return x / gcd(x, y) * y; }
// nが大きく,kが10^7程度の時の二項係数
long long com[200005];
void combination(long long n,long long k){
    com[0] = 1;
    for(int i=1;i<=k;i++){
        com[i] = (com[i-1]*(n-i+1)%MOD)*modinv(i,MOD)%MOD;
    }
}

int main(){
    int N; cin >> N;
    vector<int> A(N);
    vector<int> B(N);
    vector<int> C(N);
    for(int i=0;i<N;i++) cin >> A[i];
    for(int i=0;i<N;i++) cin >> B[i];
    for(int i=0;i<N;i++) cin >> C[i];
    sort(A.begin(),A.end());
    sort(B.begin(),B.end());
    sort(C.begin(),C.end());
    ll out=0;
    for(int i=0;i<N;i++){
        ll cnt=0;
        int l=0,r=N;
        int mid;
        cerr << B[i] << ":\t";
        while(r-l>1){
            mid = (l+r)/2;
            if(B[i]<=A[mid]){
                r=mid;
            }else{
                l=mid;
            }
        }
        cnt = l+1;
        cerr << cnt << "\t";
        l=0;r=N;
        while(r-l>1){
            mid = (l+r)/2;
            if(B[i]<C[mid]){
                r=mid;
            }else{
                l=mid;
            }
        }
        if(B[i]<C[0]) cnt *= N;
        else if(B[i]>=C[N-1]) cnt = 0;
        else cnt *= (N-r);
        cerr << (N-r) << "\t" << cnt << "\n";
        out += cnt;
    }
    cout << out << endl;
    return 0;
}