Problem: 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.

Examples

Input

3
-1 3
2 1
3 -2


Output

13


Input

4
1 4
2 1
3 3
4 2


Output

34


Input

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


Output

7222

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
#define a first
#define b second
#define sz(x) (ll)((x).size())
#define pb push_back
#define mp make_pair 
#define bg begin()
#define ed end()
#define all(x) (x).bg,(x).ed
#define rep(i,n) for(ll i=0;i<(n);i++)
#define rep1(i,n) for(ll i=1;i<=(n);i++)
#define rrep1(i,n) for(ll i=(n);i>=1;i--)
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
const ll MOD=998244353;
ll power(ll x,ll n,ll mod){ll ret=1;while(n>0){if(n&1) (ret*=x)%=mod;(x*=x)%=mod;n>>=1;}return ret;}

const ll MAX_N=265000;

struct BIT{
	ll bit[MAX_N],n;

	BIT(ll N){
		n=N;
	}

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

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

int main(){
	ll N; cin>>N;
	BIT left(N);
	BIT right(N);
	rep1(i,N) right.add(i,1);
	vector<P> p(N);
	map<ll,ll> m;
	rep(i,N){
		ll x,y; cin>>x>>y;
		p[i]=P(x,y);
		m[y]=0;
	}
	ll sub=1;
	for(auto it=m.bg;it!=m.ed;it++) it->b=sub++;

	rep(i,N) p[i].b=m[p[i].b];
	sort(all(p));

	ll ans=N*power(2,N-1,MOD);
	ans%=MOD;
	rep(i,N){
		// p[i]を見る
		// 見る前にrightを減らす
		right.add(p[i].b,-1);

		// 見る
		/*
				A|B
		 p[i].b ---
				D|C
		*/
		ll A=left.sum(p[i].b-1),B=right.sum(p[i].b-1),C=right.sum(N)-right.sum(p[i].b),D=left.sum(N)-left.sum(p[i].b);

		ll A2=power(2,A,MOD)-1,B2=power(2,B,MOD)-1,C2=power(2,C,MOD)-1,D2=power(2,D,MOD)-1;
		ans+=(((A2*C2)%MOD)*power(2,B+D,MOD))%MOD;
		ans%=MOD;
		ans+=(((B2*D2)%MOD)*power(2,A+C,MOD))%MOD;
		ans%=MOD;
		ll tmp=(A2*B2)%MOD;
		tmp*=C2;
		tmp%=MOD;
		tmp*=D2;
		tmp%=MOD;
		ans-=tmp;
		if(ans<0) ans+=MOD;
		ans%=MOD;

		// 見た後にleftを増やす
		left.add(p[i].b,1);
	}
	cout<<ans;
}