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
#include<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
struct Node
{
	int val;
	int id;
	int aorb;
}a[200005];
int n,type[200005],c[4];
bool cmp(Node a,Node b)
{
	return a.val<b.val;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		a[2*i-1].val=x;
		a[2*i-1].aorb=1;
		a[2*i-1].id=i;
		a[2*i].val=y;
		a[2*i].aorb=2;
		a[2*i].id=i;
	}
	sort(a+1,a+2*n+1,cmp);
	ll sum=0;
	for(int i=1;i<=n;i++)
	{
		sum+=a[i].val;
		c[type[a[i].id]]--;
		c[type[a[i].id]^=a[i].aorb]++;
	}
	if(c[3]||!c[1]||!c[2])
	{
		printf("%lld\n",sum);
		return 0;
	}
	sum+=a[n+1].val-a[n].val;
	c[type[a[n].id]]--;
	c[type[a[n].id]^=a[n].aorb]++;
	c[type[a[n+1].id]]--;
	c[type[a[n+1].id]^=a[n+1].aorb]++;
	if(c[3]||!c[1]||!c[2])
	{
		printf("%lld\n",sum);
		return 0;
	}
	printf("%lld\n",sum+min(a[n].val-a[n-1].val,a[n+2].val-a[n+1].val));
	return 0;
}