#include<cstdio>
#include<algorithm>
using namespace std;
#define N 105000
int n,a[N],b[N],v[N*2],id1[N],id2[N];
long long su1,su2,su[N*2],as=1e18;
long long solve(int i)
{
	if(id1[i]>n&&id2[i]>n)return su[n];
	else if(id1[i]>n+1||id2[i]>n+1)return su[n+1]-(id1[i]<=n+1?v[id1[i]]:0)-(id2[i]<=n+1?v[id2[i]]:0);
	else return su[n+2]-v[id1[i]]-v[id2[i]];
}
int main()
{
	scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]),v[i*2-1]=a[i],v[i*2]=b[i],su1+=a[i],su2+=b[i];
	sort(v+1,v+n*2+1);for(int i=1;i<=n;i++)id1[i]=lower_bound(v+1,v+n*2+1,a[i])-v,id2[i]=lower_bound(v+1,v+n*2+1,b[i])-v;
	for(int i=1;i<=n*2;i++)su[i]=su[i-1]+v[i];
	as=min(su1,su2);for(int i=1;i<=n;i++)as=min(as,solve(i));
	printf("%lld\n",as);
}//
### 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<cstdio>
#include<algorithm>
using namespace std;
#define N 105000
int n,a[N],b[N],v[N*2],id1[N],id2[N];
long long su1,su2,su[N*2],as=1e18;
long long solve(int i)
{
	if(id1[i]>n&&id2[i]>n)return su[n];
	else if(id1[i]>n+1||id2[i]>n+1)return su[n+1]-(id1[i]<=n+1?v[id1[i]]:0)-(id2[i]<=n+1?v[id2[i]]:0);
	else return su[n+2]-v[id1[i]]-v[id2[i]];
}
int main()
{
	scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]),v[i*2-1]=a[i],v[i*2]=b[i],su1+=a[i],su2+=b[i];
	sort(v+1,v+n*2+1);for(int i=1;i<=n;i++)id1[i]=lower_bound(v+1,v+n*2+1,a[i])-v,id2[i]=lower_bound(v+1,v+n*2+1,b[i])-v;
	for(int i=1;i<=n*2;i++)su[i]=su[i-1]+v[i];
	as=min(su1,su2);for(int i=1;i<=n;i++)as=min(as,solve(i));
	printf("%lld\n",as);
}//
```