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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * for (int i = 0; i < N; i++) {}
 * System.out.println();
 * HashMap<Integer, Integer> hm = new HashMap<>();
 * int[] s = new int[M];
 * ArrayDeque<Integer> deq = new ArrayDeque<>();
 *
 */
public class Main implements Runnable { //クラス名はSnukeFestival

	public static void main(String[] args) {
		Thread.setDefaultUncaughtExceptionHandler((t, e) -> System.exit(1));
		new Thread(null, new Main(), "", 16 * 1024 * 1024).start(); //16MBスタックを確保して実行
	}

	public void run() {
		FastScannerSnukeFestival sc = new FastScannerSnukeFestival(System.in);

		try {
			//String S = sc.nextLine().trim();
			int N = sc.nextInt();
			double[] a = new double[N];
			double[] b = new double[N];
			double[] c = new double[N];
			//long l = sc.nextLong();
			//ArrayList<Integer> arr = new ArrayList<Integer>();

			for (int i = 0; i < N; i++) {
				a[i] = sc.nextLong();
			}

			for (int i = 0; i < N; i++) {
				b[i] = sc.nextLong();
			}

			for (int i = 0; i < N; i++) {
				c[i] = sc.nextLong();
			}

			Arrays.sort(a);
			Arrays.sort(c);

			long ans = 0;
			long numa = 0;
			long numc = 0;
			for (int i = 0; i < N; i++) {
				/**
				 * binarySearchメソッドの戻り値は、
				 * listの中に該当する値があるときはその値の格納場所を返す。indexは0から始まる。
				 *
				 * 該当する値がないときは、その値が格納されるであろう場所の符号をマイナスにし、その値にさらに1を引いた値が返される。
				 * 符号をマイナスにするだけであると先頭に格納されたとき(index=0)は該当する値があるときとないときでは区別できなくなってしまうため、1を引く必要がある。
				 */
				//System.out.println(b[i]);
				//System.out.println(Arrays.binarySearch(a, b[i] - 0.1));
				numa = Arrays.binarySearch(a, b[i] - 0.1);
				//System.out.println(Arrays.binarySearch(c, b[i] + 0.1));
				numc = Arrays.binarySearch(c, b[i] + 0.1);

				numa = 0 - (numa + 1);
				numc = 0 - (numc + 1);

				//System.out.println(numa + " " + numc);

				if (numa > 0 && numc < N) {
					ans += numa * (N - numc);
				}
				//System.out.println("ans " + ans);
			}

			System.out.println(ans);

		} catch (Exception e) {
			e.printStackTrace();
		}

	}
}

//高速なScanner
class FastScannerSnukeFestival {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScannerSnukeFestival(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}