Problem:
Score : 600 points

Problem Statement

We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.

For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:

f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)

Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.

Constraints

1 \leq N \leq 2 \times 10^5

-10^9 \leq x_i, y_i \leq 10^9

x_i \neq x_j (i \neq j)

y_i \neq y_j (i \neq j)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.

Sample Input 1

3
-1 3
2 1
3 -2

Sample Output 1

13

Let the first, second, and third points be P_1, P_2, and P_3, respectively. S = \{P_1, P_2, P_3\} has seven non-empty subsets, and f has the following values for each of them:

f(\{P_1\}) = 1

f(\{P_2\}) = 1

f(\{P_3\}) = 1

f(\{P_1, P_2\}) = 2

f(\{P_2, P_3\}) = 2

f(\{P_3, P_1\}) = 3

f(\{P_1, P_2, P_3\}) = 3

The sum of these is 13.

Sample Input 2

4
1 4
2 1
3 3
4 2

Sample Output 2

34

Sample Input 3

10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6

Sample Output 3

7222

Be sure to print the sum modulo 998244353.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;

public class Main {
	static int mod = 998244353;
	static int DX[] = { -1, 0, 1, 0 }, DY[] = { 0, -1, 0, 1 };

	static long memo[];
	
	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		int n = fs.nextInt();
		int p[][] = new int[n][2];
		for(int i=0;i<n;i++) {
			p[i][0] = fs.nextInt();
			p[i][1] = fs.nextInt();
		}
		//座標圧縮
		//x方向
		TreeMap<Integer,Integer> tm = new TreeMap<>();
		for(int i=0;i<n;i++)tm.put(p[i][0], 0);
		int cnt = 0;
		for(int key:tm.keySet()) {
			 tm.put(key, cnt++);
		}
		for(int i=0;i<n;i++) {
			p[i][0] = tm.get(p[i][0]);
		}
		//y方向
		tm = new TreeMap<>();
		for(int i=0;i<n;i++)tm.put(p[i][1], 0);
		cnt = 0;
		for(int key:tm.keySet()) {
			 tm.put(key, cnt++);
		}
		for(int i=0;i<n;i++) {
			p[i][1] = tm.get(p[i][1]);
		}
		
		long ans = 0;

		BIT bit = new BIT(n);
		Arrays.sort(p, (p1,p2)->p1[0]-p2[0]);
		int a[] = new int[n];
		int b[] = new int[n];
		for(int i=0;i<n;i++) {
			a[i] = bit.sum(p[i][1]+1);
			b[i] = i - a[i];
			bit.add(p[i][1]+1,1);//0は入れられないので全てインクリメントする
		}
		bit = new BIT(n);
		int c[] = new int[n];
		int d[] = new int[n];
		for(int i=n-1;i>=0;i--) {
			c[i] = bit.sum(p[i][1]+1);
			d[i] = n-1-i - c[i];
			bit.add(p[i][1]+1, 1);
		}
		
		memo = new long[n+5];
		memo[0]=1;
		for(int i=1;i<n+2;i++) {
			memo[i] = mul(memo[i-1],2);
		}
		
		//左下かつ右上｜｜左上かつ右下
		for(int i=0;i<n;i++) {
			long s = mul(memo[b[i]]-1,memo[c[i]]-1);
			s = mul(s,memo[a[i]+d[i]]);
			long t = mul(memo[a[i]]-1,memo[d[i]]-1);
			t = mul(t,memo[b[i]+c[i]]);
			long st = mul(memo[a[i]]-1,memo[d[i]]-1);
			st = mul(st, memo[b[i]]-1);
			st = mul(st, memo[c[i]]-1);
			
			long res = sub(plus(s,t),st);
			ans = plus(ans,res);
			ans = plus(ans, modpow(2,n-1));
		}
		
		System.out.println(ans);
		
		
	}
	
	//MOD culculations
	static long plus(long x, long y) {
		x %= mod;
		y %= mod;
		long res = (x + y) % mod;
		return res;
	}

	static long sub(long x, long y) {
		x %= mod;
		y %= mod;
		long res = (x - y + mod) % mod;
		return res;
	}

	static long mul(long x, long y) {
		x %= mod;
		y %= mod;
		long res = x * y % mod;
		return res;
	}

	static long div(long x, long y) {
		x %= mod;
		y %= mod;
		long res = x * modpow(y, mod - 2) % mod;
		return res;
	}

	//modの元での高速pow()
	static long modpow(long a, long p) {
		if (p == 0)
			return 1;
		if (p % 2 == 0) {
			long halfP = p / 2;
			long root = modpow(a, halfP);
			return root * root % mod;
		} else
			return a * modpow(a, p - 1) % mod;
	}

	
}

//BIT Tree
class BIT {
	int n;
	long bit[];

	public BIT(int n) {
		this.n = n;
		bit = new long[n + 1];
	}

	int sum(int i) {
		int s = 0;
		while (i > 0) {
			s += bit[i];
			i -= i & -i;
		}
		return s;
	}

	void add(int i, long x) {
		while (i <= n) {
			bit[i] += x;
			i += i & -i;
		}
	}
}

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

	public FastScanner(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;
	}
}