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 Rust (1.15.1) (judge verdict: Wrong Answer)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let n = read::<usize>();
    let big_int = 998244353i64;
    let mut points = vec![];
    for i in 0..n {
        let v = read_vec::<i64>();
        points.push((v[0], v[1]));
    }
    points.sort();
    let ys = points.iter().map(|p| p.1).collect::<Vec<_>>();
    let ys = argsort(&ys);
    debug!(ys);

    let mut seg = SegTree::new(n, 0i64, |a, b| a + b);

    let mut x_smaller_y_smaller = vec![0; n];
    let mut x_smaller_y_larger = vec![0; n];
    for i in 0..n {
        x_smaller_y_smaller[i] = seg.query(0, ys[i]);
        x_smaller_y_larger[i] = i as i64 - x_smaller_y_smaller[i];
        seg.update(ys[i], 1);
    }

    let mut seg = SegTree::new(n, 0i64, |a, b| a + b);
    let mut x_larger_y_smaller = vec![0; n];
    let mut x_larger_y_larger = vec![0; n];
    for i in (0..n).rev() {
        x_larger_y_smaller[i] = seg.query(0, ys[i]);
        x_larger_y_larger[i] = (n - 1 - i) as i64 - x_larger_y_smaller[i];
        seg.update(ys[i], 1);
    }

    let mut ans = 0;
    for i in 0..n {
        assert_eq!(
            x_smaller_y_smaller[i]
                + x_larger_y_larger[i]
                + x_smaller_y_larger[i]
                + x_larger_y_smaller[i],
            n as i64 - 1
        );
        let ox1 = pow_m(2, x_smaller_y_smaller[i], big_int);
        let o1 = ox1 - 1;
        let ox2 = pow_m(2, x_larger_y_larger[i], big_int);
        let o2 = ox2 - 1;
        let ox3 = pow_m(2, x_smaller_y_larger[i], big_int);
        let o3 = ox3 - 1;
        let ox4 = pow_m(2, x_larger_y_smaller[i], big_int);
        let o4 = ox4 - 1;
        ans += ((o1 * o2) % big_int * (ox3 * ox4) % big_int) % big_int;
        ans %= big_int;
        ans += ((ox1 * ox2) % big_int * (o3 * o4) % big_int) % big_int;
        ans %= big_int;
        ans -= ((o1 * o2) % big_int * (o3 * o4) % big_int) % big_int;
        ans %= big_int;
        // debug!(ox1, ox2, ox3, ox4);
        ans += pow_m(2, n as i64 - 1, big_int);
        ans %= big_int;
        // debug!(ans);
    }
    println!("{}", (ans % big_int + big_int) % big_int);
}

fn pow_m(x: i64, p: i64, m: i64) -> i64 {
    if p == 0 {
        1
    } else if p % 2 == 0 {
        let t = pow_m(x, p / 2, m);
        t * t % m
    } else {
        let t = pow_m(x, p / 2, m);
        t * t % m * x % m
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn argsort<T: std::cmp::Ord + std::marker::Copy>(v: &Vec<T>) -> Vec<usize> {
    let mut sort_v = Vec::new();
    for i in 0..v.len() {
        sort_v.push((v[i], i));
    }
    sort_v.sort();
    sort_v.iter().map(|x| x.1).collect::<Vec<usize>>()
}

struct SegTree<I, BiOp> {
    n: usize,
    dat: Vec<I>,
    op: BiOp,
    e: I,
}

impl<I, BiOp> SegTree<I, BiOp>
where
    BiOp: Fn(I, I) -> I,
    I: Copy,
{
    pub fn new(n_: usize, e: I, op: BiOp) -> Self {
        let mut n = 1;
        while n < n_ {
            n *= 2;
        } // n is a power of 2
        SegTree {
            n: n,
            dat: vec![e; 2 * n - 1],
            op: op,
            e: e,
        }
    }
    /* ary[k] <- v */
    pub fn update(&mut self, idx: usize, v: I) {
        let mut k = idx + self.n - 1;
        self.dat[k] = v;
        while k > 0 {
            k = (k - 1) / 2;
            self.dat[k] = (self.op)(self.dat[2 * k + 1], self.dat[2 * k + 2]);
        }
    }
    /* [a, b) (note: half-inclusive)
     * http://proc-cpuinfo.fixstars.com/2017/07/optimize-segment-tree/ */
    pub fn query(&self, mut a: usize, mut b: usize) -> I {
        let mut left = self.e;
        let mut right = self.e;
        a += self.n - 1;
        b += self.n - 1;
        while a < b {
            if (a & 1) == 0 {
                left = (self.op)(left, self.dat[a]);
            }
            if (b & 1) == 0 {
                right = (self.op)(self.dat[b - 1], right);
            }
            a = a / 2;
            b = (b - 1) / 2;
        }
        (self.op)(left, right)
    }
}