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

typedef long long ll;
typedef pair<int, int> pii;

const int MAXN = 100000;

bool tag[MAXN + 5]; pii p[2*MAXN + 5]; int N;
int main() {
	ll SA = 0, SB = 0; scanf("%d", &N);
	for(int i=1;i<=N;i++) {
		int A, B; scanf("%d%d", &A, &B);
		SA += A, SB += B;
		p[2*i-1] = make_pair(A, i);
		p[2*i] = make_pair(B, i);
	}
	sort(p + 1, p + 2*N + 1);
	
	ll ans = 0; bool flag = false;
	for(int i=1;i<=N;i++) {
		if( tag[p[i].second] ) flag = true;
		else tag[p[i].second] = true;
		ans += p[i].first;
	}
	if( !flag ) {
		if( p[N].second == p[N + 1].second )
			ans = min(ans - p[N - 1].first + p[N + 1].first, ans - p[N].first + p[N + 2].first);
		else ans = ans - p[N].first + p[N + 1].first;
	}
	
	printf("%lld\n", min(ans, min(SA, SB)));
}
### Prompt

Generate a CPP solution to the following problem:

We have a directed weighted graph with N vertices. Each vertex has two integers written on it, and the integers written on Vertex i are A_i and B_i.

In this graph, there is an edge from Vertex x to Vertex y for all pairs 1 \leq x,y \leq N, and its weight is {\rm min}(A_x,B_y).

We will consider a directed cycle in this graph that visits every vertex exactly once. Find the minimum total weight of the edges in such a cycle.

Constraints

* 2 \leq N \leq 10^5
* 1 \leq A_i \leq 10^9
* 1 \leq B_i \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print the minimum total weight of the edges in such a cycle.

Examples

Input

3
1 5
4 2
6 3


Output

7


Input

4
1 5
2 6
3 7
4 8


Output

10


Input

6
19 92
64 64
78 48
57 33
73 6
95 73


Output

227

### Response

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

typedef long long ll;
typedef pair<int, int> pii;

const int MAXN = 100000;

bool tag[MAXN + 5]; pii p[2*MAXN + 5]; int N;
int main() {
	ll SA = 0, SB = 0; scanf("%d", &N);
	for(int i=1;i<=N;i++) {
		int A, B; scanf("%d%d", &A, &B);
		SA += A, SB += B;
		p[2*i-1] = make_pair(A, i);
		p[2*i] = make_pair(B, i);
	}
	sort(p + 1, p + 2*N + 1);
	
	ll ans = 0; bool flag = false;
	for(int i=1;i<=N;i++) {
		if( tag[p[i].second] ) flag = true;
		else tag[p[i].second] = true;
		ans += p[i].first;
	}
	if( !flag ) {
		if( p[N].second == p[N + 1].second )
			ans = min(ans - p[N - 1].first + p[N + 1].first, ans - p[N].first + p[N + 2].first);
		else ans = ans - p[N].first + p[N + 1].first;
	}
	
	printf("%lld\n", min(ans, min(SA, SB)));
}
```