#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL N=100010;
LL n, a[N], b[N], s[N<<1],f[N<<1];
int main(){
	scanf("%lld",&n);
	for(LL i=1; i<=n; ++i)scanf("%lld%lld",&a[i],&b[i]),s[i*2-1]=a[i],s[i*2]=b[i];
	sort(s+1,s+2*n+1);
	for(LL i=1; i<=n; ++i)f[i]=f[i-1]+s[i];
	LL ans=0, tmp=0;
	for(LL i=1; i<=n; ++i)ans+=a[i];
	for(LL i=1; i<=n; ++i)tmp+=b[i]; ans=min(ans,tmp);
	for(LL i=1; i<=n; ++i){
		tmp=a[i]+b[i]+f[n-2];
		LL d=n-2;
		if(a[i]>b[i])swap(a[i],b[i]);
		if(s[d]>=a[i])++d,tmp=tmp-a[i]+s[d];
		if(s[d]>=b[i])++d,tmp=tmp-b[i]+s[d];
		ans=min(ans,tmp);
	}
	printf("%lld",ans);
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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;
const LL N=100010;
LL n, a[N], b[N], s[N<<1],f[N<<1];
int main(){
	scanf("%lld",&n);
	for(LL i=1; i<=n; ++i)scanf("%lld%lld",&a[i],&b[i]),s[i*2-1]=a[i],s[i*2]=b[i];
	sort(s+1,s+2*n+1);
	for(LL i=1; i<=n; ++i)f[i]=f[i-1]+s[i];
	LL ans=0, tmp=0;
	for(LL i=1; i<=n; ++i)ans+=a[i];
	for(LL i=1; i<=n; ++i)tmp+=b[i]; ans=min(ans,tmp);
	for(LL i=1; i<=n; ++i){
		tmp=a[i]+b[i]+f[n-2];
		LL d=n-2;
		if(a[i]>b[i])swap(a[i],b[i]);
		if(s[d]>=a[i])++d,tmp=tmp-a[i]+s[d];
		if(s[d]>=b[i])++d,tmp=tmp-b[i]+s[d];
		ans=min(ans,tmp);
	}
	printf("%lld",ans);
	return 0;
}
```