#include <bits/stdc++.h>
#define FR first
#define SE second

using namespace std;

typedef long long ll;
typedef pair<int,int> pr;

pr a[200005],num[100005];
bool vis[100005];

int main() {
  int n;
  scanf("%d",&n);
  ll s1=0,s2=0;
  for(int i=1;i<=n;i++) {
  	int x,y;
  	scanf("%d%d",&x,&y);
  	a[2*i-1]=pr(x,i);
  	a[2*i]=pr(y,i);
  	s1+=x;s2+=y;
  	num[i]=pr(x,y);
  }
  sort(a+1,a+2*n+1);
  ll s=0;
  bool ok=0;
  for(int i=1;i<=n;i++) {
  	s+=a[i].FR;
    if (!vis[a[i].SE]) vis[a[i].SE]=1;
    else ok=1;
  }
  if (ok) {
  	printf("%lld\n",s);
  	return 0;
  }
  ll ans=min(s1,s2);
  for(int i=1;i<=n;i++) {
  	int x=a[i].SE;
  	ans=min(ans,s-a[i].FR+num[x].FR+num[x].SE-((i==n)?a[n-1].FR:a[n].FR));
  }
  printf("%lld\n",ans);
  return 0;
}
### Prompt

Please create a solution in cpp 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 FR first
#define SE second

using namespace std;

typedef long long ll;
typedef pair<int,int> pr;

pr a[200005],num[100005];
bool vis[100005];

int main() {
  int n;
  scanf("%d",&n);
  ll s1=0,s2=0;
  for(int i=1;i<=n;i++) {
  	int x,y;
  	scanf("%d%d",&x,&y);
  	a[2*i-1]=pr(x,i);
  	a[2*i]=pr(y,i);
  	s1+=x;s2+=y;
  	num[i]=pr(x,y);
  }
  sort(a+1,a+2*n+1);
  ll s=0;
  bool ok=0;
  for(int i=1;i<=n;i++) {
  	s+=a[i].FR;
    if (!vis[a[i].SE]) vis[a[i].SE]=1;
    else ok=1;
  }
  if (ok) {
  	printf("%lld\n",s);
  	return 0;
  }
  ll ans=min(s1,s2);
  for(int i=1;i<=n;i++) {
  	int x=a[i].SE;
  	ans=min(ans,s-a[i].FR+num[x].FR+num[x].SE-((i==n)?a[n-1].FR:a[n].FR));
  }
  printf("%lld\n",ans);
  return 0;
}
```