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

#define maxn 100010 

int n ;
ll a[maxn],b[maxn] ;
ll c[maxn*2],presum[maxn*2] ;
ll sum, ans, suma, sumb ;

int main(void) {
	//freopen("1.txt","r",stdin) ;

	scanf("%d",&n) ;
	for(int i=0; i<n; i++) {
		scanf("%lld%lld", &a[i], &b[i]) ;
		suma += a[i] ;
		sumb += b[i] ;	
	}
	ans = min(suma, sumb ) ;
	for(int i=0; i<n; i++ ) {
		c[i*2] = a[i] ;
		c[i*2+1] = b[i] ;
	}
	sort(c,c+2*n) ;
	presum[0] = c[0] ;
	for(int i=1; i<=2*n; i++) {
		presum[i] = presum[i-1]+c[i] ;
	}
	for(int i=0; i<n; i++) {
		int pos = n - 2 ;
		int tmp = 0 ;
		if( a[i]>c[pos] ) tmp += a[i] ;
		else pos ++ ;
		if( b[i]>c[pos] ) tmp += b[i] ;
		else pos ++ ;
		ans = min( ans, tmp + presum[pos-1]) ;
	}
	printf("%lld\n",ans ) ;
	return 0 ;
}
### Prompt

Create a solution in cpp for 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 <bits/stdc++.h>
 
using namespace std;
typedef long long ll ;

#define maxn 100010 

int n ;
ll a[maxn],b[maxn] ;
ll c[maxn*2],presum[maxn*2] ;
ll sum, ans, suma, sumb ;

int main(void) {
	//freopen("1.txt","r",stdin) ;

	scanf("%d",&n) ;
	for(int i=0; i<n; i++) {
		scanf("%lld%lld", &a[i], &b[i]) ;
		suma += a[i] ;
		sumb += b[i] ;	
	}
	ans = min(suma, sumb ) ;
	for(int i=0; i<n; i++ ) {
		c[i*2] = a[i] ;
		c[i*2+1] = b[i] ;
	}
	sort(c,c+2*n) ;
	presum[0] = c[0] ;
	for(int i=1; i<=2*n; i++) {
		presum[i] = presum[i-1]+c[i] ;
	}
	for(int i=0; i<n; i++) {
		int pos = n - 2 ;
		int tmp = 0 ;
		if( a[i]>c[pos] ) tmp += a[i] ;
		else pos ++ ;
		if( b[i]>c[pos] ) tmp += b[i] ;
		else pos ++ ;
		ans = min( ans, tmp + presum[pos-1]) ;
	}
	printf("%lld\n",ans ) ;
	return 0 ;
}
```