#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

typedef long long ll;
typedef pair<int,int> P;
const ll MOD = 998244353;

int main(){
	static int n;
	static int a[100010];
	static int b[100010];
	scanf("%d",&n);
	for(int i = 1 ; i <= n ; i ++)scanf("%d",&a[i]);
	for(int i = 1 ; i <= n ; i ++)scanf("%d",&b[i]);
	
	vector<P> A;
	for(int i = 1 ; i <= n ; i ++){
		A.push_back(P(a[i],i));
	}
	sort(A.begin(),A.end());
	vector<P> B;
	for(int i = 1 ; i <= n ; i ++){
		B.push_back(P(b[i],i));
	}
	sort(B.begin(),B.end());
	
	static int nex[100010];
	for(int i = 1 ; i <= n ; i ++){
		nex[A[i-1].second] = B[i-1].second;
	}
	for(int i = 1 ; i <= n ; i ++){
		if(A[i-1].first > B[i-1].first){
			puts("No");
			return 0;
		}
	}
	
	bool ret = false;
	for(int i = 1 ; i+1 <= n ; i ++){
		if(A[i].first <= B[i-1].first)ret = true;
	}
	int cnt = 1, loc = nex[1];
	while(loc != 1){
		cnt ++;
		loc = nex[loc];
	}
	if(cnt != n)ret = true;
	
	if(ret)puts("Yes");
	else puts("No");
}
### Prompt

Create a solution in cpp for the following problem:

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`.

Examples

Input

3
1 3 2
1 2 3


Output

Yes


Input

3
1 2 3
2 2 2


Output

No


Input

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


Output

Yes

### Response

```cpp
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

typedef long long ll;
typedef pair<int,int> P;
const ll MOD = 998244353;

int main(){
	static int n;
	static int a[100010];
	static int b[100010];
	scanf("%d",&n);
	for(int i = 1 ; i <= n ; i ++)scanf("%d",&a[i]);
	for(int i = 1 ; i <= n ; i ++)scanf("%d",&b[i]);
	
	vector<P> A;
	for(int i = 1 ; i <= n ; i ++){
		A.push_back(P(a[i],i));
	}
	sort(A.begin(),A.end());
	vector<P> B;
	for(int i = 1 ; i <= n ; i ++){
		B.push_back(P(b[i],i));
	}
	sort(B.begin(),B.end());
	
	static int nex[100010];
	for(int i = 1 ; i <= n ; i ++){
		nex[A[i-1].second] = B[i-1].second;
	}
	for(int i = 1 ; i <= n ; i ++){
		if(A[i-1].first > B[i-1].first){
			puts("No");
			return 0;
		}
	}
	
	bool ret = false;
	for(int i = 1 ; i+1 <= n ; i ++){
		if(A[i].first <= B[i-1].first)ret = true;
	}
	int cnt = 1, loc = nex[1];
	while(loc != 1){
		cnt ++;
		loc = nex[loc];
	}
	if(cnt != n)ret = true;
	
	if(ret)puts("Yes");
	else puts("No");
}
```