#include <bits/stdc++.h>
using namespace std;
const int N=200010;
int n,a[N],b[N];
long long ans=LLONG_MAX,sum1,sum2,sum[N];
int main(){
    scanf("%d",&n);
    for (int i=1; i<=n; ++i) scanf("%d%d",&a[i],&b[i]);
    long long sum1=0,sum2=0;
    multiset<int> s;
    for (int i=1; i<=n; ++i){
	sum1+=a[i];
	sum2+=b[i];
	s.insert(a[i]);
	s.insert(b[i]);
    }
    for (int i=1; i<=n; ++i){
	sum[i]=sum[i-1]+(*s.begin());
	s.erase(s.begin());
    }
    for (int i=1; i<=n; ++i){
	int x=a[i],y=b[i];
	if (x>y) swap(x,y);
	if (x<=sum[n-1]-sum[n-2]){
	    if (y<=sum[n]-sum[n-1]) ans=min(ans,sum[n]);
	    else ans=min(ans,sum[n-1]+y);
	}
	else ans=min(ans,sum[n-2]+x+y);
    }
    cout<<min(min(ans,sum1),sum2);
}
### 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>
using namespace std;
const int N=200010;
int n,a[N],b[N];
long long ans=LLONG_MAX,sum1,sum2,sum[N];
int main(){
    scanf("%d",&n);
    for (int i=1; i<=n; ++i) scanf("%d%d",&a[i],&b[i]);
    long long sum1=0,sum2=0;
    multiset<int> s;
    for (int i=1; i<=n; ++i){
	sum1+=a[i];
	sum2+=b[i];
	s.insert(a[i]);
	s.insert(b[i]);
    }
    for (int i=1; i<=n; ++i){
	sum[i]=sum[i-1]+(*s.begin());
	s.erase(s.begin());
    }
    for (int i=1; i<=n; ++i){
	int x=a[i],y=b[i];
	if (x>y) swap(x,y);
	if (x<=sum[n-1]-sum[n-2]){
	    if (y<=sum[n]-sum[n-1]) ans=min(ans,sum[n]);
	    else ans=min(ans,sum[n-1]+y);
	}
	else ans=min(ans,sum[n-2]+x+y);
    }
    cout<<min(min(ans,sum1),sum2);
}
```