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.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class Main {
	
	static final long mod = 998244353L;

	public static void main(String[] args) {
		Main mainObj = new Main();
		mainObj.solve();
	}
	
	void solve() {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		
		ArrayList<Coodinate> coodinateList = new ArrayList<>();
		for(int i = 0; i < n; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			coodinateList.add(new Coodinate(x,y,i));
		}
		
		sc.close();
		
		Comparator<Coodinate> yComparator = Comparator.comparing(Coodinate::getY);
		Comparator<Coodinate> xComparator = Comparator.comparing(Coodinate::getX);
		coodinateList.sort(yComparator);
		for(int i = 0; i < n; i++) {
			coodinateList.get(i).setY(i);
		}
		
		coodinateList.sort(xComparator);
		for(int i = 0; i < n; i++) {
			coodinateList.get(i).setX(i);
		}
		
		int[] firstQuadrant = new int[n];
		int[] secondQuadrant = new int[n];
		int[] thirdQuadrant = new int[n];
		int[] forthQuadrant = new int[n];
		
		// 平面走査する
		SumSegTree leftTree = new SumSegTree(n);
		// xの小さい方
		for(int i = 0; i < n; i++) {
			
			Coodinate nowCoodinate = coodinateList.get(i);
			
			int numOfThirdQuadrant = leftTree.getValue(0, nowCoodinate.getY(), 0, 0, -1);
			thirdQuadrant[i] = numOfThirdQuadrant;
			secondQuadrant[i] = i - numOfThirdQuadrant;
			
			leftTree.upDate(nowCoodinate.getY(), 1);
		}
		
		SumSegTree rightTree = new SumSegTree(n);
		// xの大きい方
		for(int i = n - 1; i >= 0; i--) {
			
			Coodinate nowCoodinate = coodinateList.get(i);
			
			int numOfForthQuadrant = rightTree.getValue(0, nowCoodinate.getY(), 0, 0, -1);
			forthQuadrant[i] = numOfForthQuadrant;
			//System.out.println(numOfForthQuadrant);
			firstQuadrant[i] = (n - 1) - i - numOfForthQuadrant;
			rightTree.upDate(nowCoodinate.getY(), 1);
		}
		
		long ans = 0;
		long[] powArr = new long[n+1];
		powArr[0] = 1;
		powArr[1] = 1;
		for(int i = 1 ; i < n; i++) {
			powArr[i] = calcMod(powArr[i - 1]*2);
		}
		
		
		
		for(int i = 0; i < n ;i++) {
			int first = firstQuadrant[i];
			int second = secondQuadrant[i];
			int third = thirdQuadrant[i];
			int forth = forthQuadrant[i];
			
			long firstPow,secondPow,thirdPow,forthPow;
			firstPow = powArr[first];
			secondPow = powArr[second];
			thirdPow = powArr[third];
			forthPow = powArr[forth];
			
			ans = calcMod(ans + calcMod(calcMod(calcMod(firstPow*(secondPow-1))*thirdPow)*(forthPow - 1)));
			ans = calcMod(ans + calcMod(calcMod(calcMod((firstPow-1)*secondPow)*(thirdPow - 1))*forthPow));
			ans = calcMod(ans - calcMod(calcMod(calcMod((firstPow-1)*(secondPow - 1))*(thirdPow - 1))*(forthPow - 1)));
			ans = calcMod(ans + calcMod(calcMod(calcMod(firstPow*secondPow)*thirdPow)*forthPow));
		}
		
		System.out.println(ans);
	}
	
	long calcMod(long value) {
		if(value < 0) {
			while(value > 0) {
				value += mod;
			}
			return value;
		}else {
			return value%mod;	
		}
	}
	
	long powWithMod(int base, int times) {
		long retValue = 1;
		for(int i = 0;i < times; i++) {
			retValue = calcMod(retValue*base);
		}
		return retValue;
	}
	
	class SumSegTree{
		int[] arr;
		int n;
		int arrSize;
		
		SumSegTree(int n){
			this.n = n;
			arrSize = 1;
			while(n > arrSize) {
				arrSize *= 2;
			}
			
			// 最初は和もvalueも0なので何もしない
			this.arr = new int[arrSize*2 - 1];
		}
		
		void upDate(int pos, int value) {
			int segIndex = pos + arrSize - 1;
			int beforeValue = arr[segIndex];
			arr[segIndex] = value;
			
			int diff = value - beforeValue;
			while(segIndex > 0) {
				segIndex = (segIndex - 1)/2;
				arr[segIndex] = arr[segIndex] + diff;
			}
		}
		
		// a, b間の和を取得
		int getValue(int a, int b, int k, int l, int r) {
			
			if(r < 0) {
				r = arrSize;
			}
			
			if(r <= a || b <= l) {
				return 0;
			}
			if(a <= l && r <= b) {
				return arr[k];
			}
			
			return getValue(a, b, 2*k + 1, l, (l+r)/2) + getValue(a,b,2*k + 2, (l+r)/2, r);
		}
	}
	
	class Coodinate{
		int x;
		int y;
		int no;
		
		Coodinate(int x, int y, int no){
			this.x = x;
			this.y = y;
			this.no = no;
		}
		
		int getX() {
			return x;
		}
		
		int getY() {
			return y;
		}
		
		void setX(int x) {
			this.x = x;
		}
		
		void setY(int y) {
			this.y = y;
		}
		
		int getNo() {
			return no;
		}
	}
}