Problem:
Score : 600 points

Problem Statement

Given are two integer sequences of N elements each: A_1,...,A_N and B_1,...,B_N.
Determine if it is possible to do the following operation at most N-2 times (possibly zero) so that, for every integer i from 1 to N, A_i \leq B_i holds:

Choose two distinct integers x and y between 1 and N (inclusive), and swap the values of A_x and A_y.

Constraints

2 \leq N \leq 10^5

1 \leq A_i,B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

If the objective is achievable, print Yes; if it is not, print No.

Sample Input 1

3
1 3 2
1 2 3

Sample Output 1

Yes

We should swap the values of A_2 and A_3.

Sample Input 2

3
1 2 3
2 2 2

Sample Output 2

No

Sample Input 3

6
3 1 2 6 3 4
2 2 8 3 4 3

Sample Output 3

Yes

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define clr(x) memset(x,0,sizeof(x))
#define For(i,a,b) for (int i=(a);i<=(b);i++)
#define Fod(i,b,a) for (int i=(b);i>=(a);i--)
#define fi first
#define se second
#define kill _z_kill
#define y0 _z_y0
#define y1 _z_y1
#define x0 _z_x0
#define x1 _z_x1
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef unsigned uint;
typedef long double LD;
typedef vector <int> vi;
typedef pair <int,int> pii;
void enable_comma(){}
string tostring(char c);
string tostring(LL x);
template <class A,class B> string tostring(pair <A,B> p);
template <class A> string tostring(vector <A> v);
string tostring(char c){
	string s="";
	s+=c;
	return s;
}
string tostring(string s){
	return "\""+s+"\"";
}
string tostring(char *c){
	return tostring((string)c);
}
string tostring(LL x){
	if (x<0)
		return "-"+tostring(-x);
	if (x>9)
		return tostring(x/10)+tostring(char('0'+x%10));
	else
		return tostring(char('0'+x));
}
string tostring(int x){
	return tostring((LL)x);
}
string tostring(ull x){
	if (x>9)
		return tostring((LL)(x/10))+tostring(char('0'+x%10));
	else
		return tostring(char('0'+x));
}
string tostring(uint x){
	return tostring((LL)x);
}
template <class A,class B> string tostring(pair <A,B> p){
	return "("+tostring(p.fi)+","+tostring(p.se)+")";
}
template <class A> string tostring(vector <A> v){
	string res="{";
	For(i,0,(int)v.size()-1){
		res+=tostring(v[i]);
		res+=i==(int)v.size()-1?"}":",";
	}
	return res;
}
template <class A> string tostring(A *a,int L,int R){
	string res="{";
	For(i,L,R){
		res+=tostring(a[i]);
		res+=i==R?"}":",";
	}
	return res;
}
string tostrings(){
	return "";
}
template <typename Head,typename... Tail>
string tostrings(Head H,Tail... T){
	return tostring(H)+" "+tostrings(T...);
}
#define User_Time ((double)clock()/CLOCKS_PER_SEC)
#ifdef zzd
	#define outval(x) cerr<<#x" = "<<tostring(x)<<endl
	#define outvals(...) cerr<<"["<<#__VA_ARGS__<<"]: "<<tostrings(__VA_ARGS__)<<endl
	#define outtag(x) cerr<<"--------------"#x"---------------"<<endl
	#define outsign(x) cerr<<"<"#x">"<<endl
	#define outarr(a,L,R) cerr<<#a"["<<(L)<<".."<<(R)<<"] = "<<tostring(a,L,R)<<endl
#else
	#define outval(x) enable_comma()
	#define outvals(...) enable_comma()
	#define outtag(x) enable_comma()
	#define outsign(x) enable_comma()
	#define outarr(a,L,R) enable_comma()
#endif
#ifdef ONLINE_JUDGE
	#ifdef assert
		#undef assert
	#endif
	#define assert(x) (!(x)?\
			cout<<"Assertion failed!"<<endl<<\
			"function: "<<__FUNCTION__<<endl<<\
			"line: "<<__LINE__<<endl<<\
			"expression: "<<#x<<endl,exit(3),0:1)
#endif
LL read(){
	LL x=0,f=0;
	char ch=getchar();
	while (!isdigit(ch))
		f=ch=='-',ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return f?-x:x;
}
template <class T> void ckmax(T &x,const T y){
	if (x<y)
		x=y;
}
template <class T> void ckmin(T &x,const T y){
	if (x>y)
		x=y;
}
//int Pow(int x,int y){
//	int ans=1;
//	for (;y;y>>=1,x=(LL)x*x%mod)
//		if (y&1)
//			ans=(LL)ans*x%mod;
//	return ans;
//}
//void Add(int &x,int y){
//	if ((x+=y)>=mod)
//		x-=mod;
//}
//void Del(int &x,int y){
//	if ((x-=y)<0)
//		x+=mod;
//}
//int Add(int x){
//	return x>=mod?x-mod:x;
//}
//int Del(int x){
//	return x<0?x+mod:x;
//}
//int md(LL x){
//	return (x%mod+mod)%mod;
//}
const int N=100005;
int n;
int a[N],b[N];
int A[N],B[N];
int p[N];
int vis[N];
int main(){
	n=read();
	For(i,1,n)
		a[i]=read(),A[i]=i;
	For(i,1,n)
		b[i]=read(),B[i]=i;
	sort(A+1,A+n+1,[&](int x,int y){
		return a[x]<a[y];
	});
	sort(B+1,B+n+1,[&](int x,int y){
		return b[x]<b[y];
	});
	For(i,1,n)
		if (a[A[i]]>b[B[i]])
			return puts("No"),0;
	For(i,1,n-1)
		if (a[A[i+1]]<=b[B[i]])
			return puts("Yes"),0;
	For(i,1,n)
		p[A[i]]=B[i];
	for (int i=1;!vis[i];i=p[i])
		vis[i]=1;
	For(i,1,n)
		if (!vis[i])
			return puts("Yes"),0;
	puts("No");
	return 0;
}