#include<bits/stdc++.h>
#define maxn 100005
#define ll long long
using namespace std;
inline int read(){
	int x=0,c;while(!isdigit(c=getchar()));
	while(x=x*10+c-'0',isdigit(c=getchar()));
	return x;
}
ll a[maxn],b[maxn],c[maxn<<1],sum[maxn],n;
int main(){
	n=read();
	for(int i=1;i<=n;i++)a[i]=read(),b[i]=read();
	ll ans=0,tmp=0;
	for(int i=1;i<=n;i++)ans+=a[i];
	for(int i=1;i<=n;i++)tmp+=b[i];
	ans=min(ans,tmp);
	for(int i=1;i<=n;i++)c[2*i]=a[i],c[2*i-1]=b[i];
	sort(c+1,c+2*n+1);
	for(int i=1;i<=n;i++)sum[i]=sum[i-1]+c[i];
	for(int i=1;i<=n;i++){
		tmp=0;int now=n;
		if(a[i]>=c[now])tmp+=a[i],now--;
		if(b[i]>=c[now])tmp+=b[i],now--;
		ans=min(ans,tmp+sum[now]);
	}printf("%lld\n",ans);return 0;
}
### 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<bits/stdc++.h>
#define maxn 100005
#define ll long long
using namespace std;
inline int read(){
	int x=0,c;while(!isdigit(c=getchar()));
	while(x=x*10+c-'0',isdigit(c=getchar()));
	return x;
}
ll a[maxn],b[maxn],c[maxn<<1],sum[maxn],n;
int main(){
	n=read();
	for(int i=1;i<=n;i++)a[i]=read(),b[i]=read();
	ll ans=0,tmp=0;
	for(int i=1;i<=n;i++)ans+=a[i];
	for(int i=1;i<=n;i++)tmp+=b[i];
	ans=min(ans,tmp);
	for(int i=1;i<=n;i++)c[2*i]=a[i],c[2*i-1]=b[i];
	sort(c+1,c+2*n+1);
	for(int i=1;i<=n;i++)sum[i]=sum[i-1]+c[i];
	for(int i=1;i<=n;i++){
		tmp=0;int now=n;
		if(a[i]>=c[now])tmp+=a[i],now--;
		if(b[i]>=c[now])tmp+=b[i],now--;
		ans=min(ans,tmp+sum[now]);
	}printf("%lld\n",ans);return 0;
}
```