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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;

#define REP(i,n) for(long long i = 0; i < (n); i++)
#define FOR(i, m, n) for(long long i = (m);i < (n); ++i)
#define ALL(obj) (obj).begin(),(obj).end()
#define SPEED cin.tie(0);ios::sync_with_stdio(false);

template<class T> using V = vector<T>;
template<class T, class U> using P = pair<T, U>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQR = priority_queue<T,vector<T>,greater<T>>;

constexpr long long MOD = (long long)1e9 + 7;
constexpr long long MOD2 = 998244353;
constexpr long long HIGHINF = (long long)1e18;
constexpr long long LOWINF = (long long)1e15;
constexpr long double PI = 3.1415926535897932384626433;

template <class T> vector<T> make_v(size_t N,T init){return vector<T>(N,init);}
template <class... T> auto make_v(size_t N,T... t){return vector<decltype(make_v(t...))>(N,make_v(t...));}
template <class T> void corner(bool flg, T hoge) {if (flg) {cout << hoge << endl; exit(0);}}
template <class T, class U>ostream &operator<<(ostream &o, const map<T, U>&obj) {o << "{"; for (auto &x : obj) o << " {" << x.first << " : " << x.second << "}" << ","; o << " }"; return o;}
template <class T>ostream &operator<<(ostream &o, const set<T>&obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr) o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
template <class T>ostream &operator<<(ostream &o, const multiset<T>&obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr) o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
template <class T>ostream &operator<<(ostream &o, const vector<T>&obj) {o << "{"; for (int i = 0; i < (int)obj.size(); ++i)o << (i > 0 ? ", " : "") << obj[i]; o << "}"; return o;}
template <class T, class U>ostream &operator<<(ostream &o, const pair<T, U>&obj) {o << "{" << obj.first << ", " << obj.second << "}"; return o;}
template <template <class tmp>  class T, class U> ostream &operator<<(ostream &o, const T<U> &obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr)o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
void print(void) {cout << endl;}
template <class Head> void print(Head&& head) {cout << head;print();}
template <class Head, class... Tail> void print(Head&& head, Tail&&... tail) {cout << head << " ";print(forward<Tail>(tail)...);}
template <class T> void chmax(T& a, const T b){a=max(a,b);}
template <class T> void chmin(T& a, const T b){a=min(a,b);}
void YN(bool flg) {cout << (flg ? "YES" : "NO") << endl;}
void Yn(bool flg) {cout << (flg ? "Yes" : "No") << endl;}
void yn(bool flg) {cout << (flg ? "yes" : "no") << endl;}



template<class T> class Segment_Tree_Range_Sum_Query {
	size_t N,M;
	T ini;
	vector<T> node;

public:
	Segment_Tree_Range_Sum_Query(const vector<T> & ar, const T ini) : M(ar.size()), ini(ini) {
		for (N = 1; N < M; N *= 2);
		node.resize(2 * N - 1, ini);
		for (int i = 0; i < M; ++i) node[i + N - 1] = ar[i];
		for (int i = N - 2; i >= 0; --i) node[i] = node[2 * i + 1]+node[2 * i + 2];
	}
 
	Segment_Tree_Range_Sum_Query(const size_t M, const T ini) : M(M), ini(ini) {
		for (N = 1; N < M; N *= 2);
		node.resize(2 * N - 1, ini);
	}
	
	void update(size_t idx, const T var) {
		idx += (N - 1);
		node[idx] += var;
		while (idx > 0) {
			idx = (idx - 1) / 2;
			node[idx] = node[2 * idx + 1]+node[2 * idx + 2];
		}
	}
 
	T getvar(const int a, const int b, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = N;
		if (r <= a || b <= l) return ini;
		if (a <= l && r <= b) return node[k];
		T vl = getvar(a, b, 2 * k + 1, l, (l + r) / 2);
		T vr = getvar(a, b, 2 * k + 2, (l + r) / 2, r);
		return vl+vr;
	}
	
	T operator[](size_t idx) {
		return node[idx + N - 1];
	}

	T operator[](pair<size_t, size_t> p) {
		return getvar(p.first, p.second);
	}

 
	void print(){
		cout << "{ " << getvar(0,1);
		for(int i = 1; i < M; ++i) cout << ", " << getvar(i,i+1);
		cout << " }" << endl;
	}
};

//Seat Pressure O(NlogN)
template<class T> vector<int> Zarts(const vector<T> & ar, int cnt = 0) {
    vector<int> res;
    map<T, int> mp;
    for (auto &e : ar) mp[e] = 0;
    for (auto &e : mp) e.second = cnt++;
    for (auto &e : ar) res.push_back(mp[e]);
    return res;
}

int main() {
    int N; cin >> N;
    vector<int> s(N),t(N),x(N),y(N),idx(N);
    for(int i = 0; i < N; ++i) cin >> s[i] >> t[i];
    x = Zarts(s);
    y = Zarts(t);
    Segment_Tree_Range_Sum_Query<ll> Seg(N,0); 
    vector<vector<ll>> cnt(N,vector<ll>(4,0));
    iota(ALL(idx),0);
    sort(ALL(idx),[&](int l,int r){return y[l]<y[r];});
    vector<ll> pow2(N,1);
    for(int i = 1; i < N; ++i) pow2[i] = (pow2[i-1]*2LL) % MOD2;
    for(auto i:idx) {
        cnt[i][3] = Seg.getvar(x[i]+1,N);
        cnt[i][2] = Seg.getvar(0,x[i]);
        cnt[i][1] = x[i] - cnt[i][2];
        cnt[i][0] = N-1 - x[i] - cnt[i][3];
        for(int j = 0; j < 4; ++j) cnt[i][j] = (pow2[cnt[i][j]] - 1 + MOD2) %MOD2;
        Seg.update(x[i],1);
    }
    ll ans = N;
    for(auto c:cnt) {
        for(int i = 0; i < 4; ++i) (ans += c[i]) %= MOD2;
        for(int i = 0; i < 4; ++i) (ans += (c[i]*c[(i+1)%4])%MOD2 ) %= MOD2;
        for(int i = 0; i < 2; ++i) (ans += (2LL*(c[i]*c[(i+2)%4])%MOD2) %MOD2 ) %= MOD2;
        for(int i = 0; i < 4; ++i) (ans += ( ((2LL*c[i])%MOD2) * ((c[(i+1)%4]*c[(i+2)%4])%MOD2) )%MOD2) %= MOD2;
        (ans += (2LL*( ((c[0]*c[1])%MOD2) * ((c[2]*c[3])%MOD2) )%MOD2 )%MOD2 ) %= MOD2;
    }
    cout << ans << endl;

	return 0;
}