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 <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <fstream> 
#define _USE_MATH_DEFINES
#include <math.h>
#include <unordered_set>

using namespace std;


inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }

template<class T> inline T sqr(T x) { return x*x; }

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<long long int> vll;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  FOR(i,0,n)
#define P(p) cout<<(p)<<endl;
#define VEC_2D(a,b) vector<vector<int> >(a, vector<int>(b, 0))
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define pb push_back
#define mp make_pair
#define INF (1000000000)
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define MOD 1000000007LL
#define FSP(a) cout << fixed << setprecision(a)


template<typename T> T gcd(T x, T y) {
	if (y == 0) return x;
	else return gcd(y, x%y);
}

template<typename T> T lcm(T a, T b) {
	return a / gcd(a, b) * b;
}
template<typename T> bool is_prime(T n) {
	for (int i = 2; i * i <= n; i++) {
		if (n % i == 0) return false;
	}
	return n != 1;
}
map<int, int> prime_factor(int n) {
	map<int, int> res;
	for (int i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			++res[i];
			n /= i;
		}
	}
	if (n != 1) res[n] = 1;
	return res;
}
int extgcd(int a, int b, int& x, int& y) {//
	int d = a;
	if (b != 0) {
		d = extgcd(b, a%b, y, x);
		y -= (a / b)*x;
	}
	else {
		x = 1; y = 0;
	}
	return d;
}



ll mod_pow(ll x, ll n, ll mod) {
	if (n == 0) return 1;
	ll res = mod_pow(x * x % mod, n / 2, mod);
	if (n & 1) res = res * x % mod;
	return res;
}
vector<string> split(const string &str, char delim) {
	vector<string> res;
	size_t current = 0, found;
	while ((found = str.find_first_of(delim, current)) != string::npos) {
		res.push_back(string(str, current, found - current));
		current = found + 1;
	}
	res.push_back(string(str, current, str.size() - current));
	return res;
}

bool is_kadomatsu(int a, int b, int c) {
	if (a == b || a == c || b == c)return false;
	if (a > b && c > b) return true;
	if (a < b && c < b)return true;
	return false;
}


struct UF {
	int n;
	vi d;
	UF() {}
	UF(int n) :n(n), d(n, -1) {}
	int root(int v) {
		if (d[v] < 0) return v;
		return d[v] = root(d[v]);
	}
	bool same(int a, int b) {
		return root(a) == root(b);
	}
	bool unite(int x, int y) {
		x = root(x); y = root(y);
		if (x == y) return false;
		if (size(x) < size(y)) swap(x, y);
		d[x] += d[y];
		d[y] = x;
		return true;
	}
	int size(int v) { return -d[root(v)]; }
};



vector<int> divisor(int n) {
	if (n == 1) return{};
	vi res;
	for (int i = 1; i*i <= n; i++) {
		if (n%i == 0) {
			res.emplace_back(i);
			if (i != 1 && i != n / i)res.emplace_back(n / i);
		}
	}
	return res;
}

struct Bellmanford {
int n;
struct edge {
	int from, to, cost;
};
vector<edge> E;
vi d;
Bellmanford(int n) :n(n), d(n) {
	E.resize(n);
}
void add_edge(int x, int y, int cost) {
	edge e;
	e.from = x; e.to = y; e.cost = cost;
	E.push_back(e);
}
void shortest_path(int s) {
rep(i, n)d[i] = INF;
d[s] = 0;
while (true) {
	bool update = false;
	for (auto e : E) {
		if (d[e.from] != INF && d[e.to] > d[e.from] + e.cost) {
			d[e.to] = d[e.from] + e.cost;
			update = true;
		}
	}
	if (!update) break;
}
}
};
struct Dijkstra {
	int n;
	struct edge {
		int to; ll cost;
	};
	vector<vector<edge>> G;
	vll d;
	Dijkstra(int n) :n(n), d(n) {
		G.resize(n);
	}
	void add_edge(int x, int y, ll cost) {
		edge e;
		e.to = y; e.cost = cost;
		G[x].push_back(e);
	}
	void shortest_path(int s) {
		rep(i, n)d[i] = 100000000000000000;
		d[s] = 0;
		priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> que;
		que.push(make_pair(0, s));

		while (!que.empty()) {
			pii p = que.top(); que.pop();
			int v = p.second;
			if (d[v] < p.first) continue;
			for (auto e : G[v]) {
				if (d[e.to] > d[v] + e.cost) {
					d[e.to] = d[v] + e.cost;
					que.push(make_pair(d[e.to], e.to));
				}
			}
		}
	}
};

struct Segmenttree {
	int n;
	vector<pair<ll, int>> dat;
	Segmenttree() {}
	void init(ll n_) {
		n = 1;
		while (n < n_) n *= 2;
		dat.resize(2 * n - 1);
		rep(i, 2 * n - 1)dat[i] = pair<ll, int>(-INF, -INF);
	}
	void update(int idx, ll val) {
		idx += n - 1;
		dat[idx] = make_pair(val, -(idx - n + 1));
		while (idx > 0) {
			idx = (idx - 1) / 2;
			dat[idx] = max(dat[idx * 2 + 1], dat[idx * 2 + 2]);
		}
	}
	pair<ll, int> query(int a, int b) {
		return query_seg(a, b, 0, 0, n);
	}
	pair<ll, int> query_seg(int a, int b, int k, int l, int r) {
		if (r <= a || b <= l) return pair<ll, int>(-INF, -INF);
		if (a <= l && r <= b)return dat[k];
		else {
			return max(query_seg(a, b, k * 2 + 1, l, (l + r) / 2), query_seg(a, b, k * 2 + 2, (l + r) / 2, r));
		}
	}
};
//-------------------------------------------------------------

int sum(ll k) {
	int ret = 0;
	while (k > 0) {
		ret += k % 10;
		k /= 10;
	}
	return ret;
}
int main() {

	ll k;
	cin >> k;
	int ans = sum(k);
	for (ll i = 1; i <= 59999999; i++) {
		ll tmp = k*i;
		ans = min(ans, sum(tmp));
	}
	while (k % 2 == 0) {
		k *= 5;
		k /= 10;
	}
	while (k % 10 == 5) {
		k *= 2;
		k /= 10;
	}
	if (k % 10 == 1) {
		P(min(ans,sum(k)))
	}
	else if(k%10==3){
		P(min(ans, sum(k*7)))
	}
	else if (k % 10 == 7) {
		P(min(ans, sum(k*3)))
	}
	else {
		P(min(ans, sum(k*9)))
	}
	return 0;

}