#include <bits/stdc++.h>
using namespace std;

int n;
long long s[200010],ans,a[100010],b[100010],c[200010];
int main()
{
 scanf("%d",&n);
 long long ji1=0,ji2=0;
 for(int i=1;i<=n;++i)
 {
  scanf("%lld%lld",&a[i],&b[i]);
  c[i]=a[i];
  c[i+n]=b[i];
  ji1+=a[i];
  ji2+=b[i];
 }
 ans=min(ji1,ji2);
 sort(c+1,c+2*n+1);
 for(int i=1;i<=n;++i)
 s[i]=s[i-1]+c[i];
 for(int i=1;i<=n;++i)
 {
  long long x=a[i],y=b[i];
  if(x>y)
  swap(x,y);
  if(lower_bound(c+1,c+2*n+1,x)-c<=n-2)
  {
   if(lower_bound(c+1,c+2*n+1,y)-c<=n-1)
   ans=min(ans,s[n]);
   else
   ans=min(ans,y+s[n-1]);
  }
  else
  ans=min(ans,x+y+s[n-2]);
 }
 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>
using namespace std;

int n;
long long s[200010],ans,a[100010],b[100010],c[200010];
int main()
{
 scanf("%d",&n);
 long long ji1=0,ji2=0;
 for(int i=1;i<=n;++i)
 {
  scanf("%lld%lld",&a[i],&b[i]);
  c[i]=a[i];
  c[i+n]=b[i];
  ji1+=a[i];
  ji2+=b[i];
 }
 ans=min(ji1,ji2);
 sort(c+1,c+2*n+1);
 for(int i=1;i<=n;++i)
 s[i]=s[i-1]+c[i];
 for(int i=1;i<=n;++i)
 {
  long long x=a[i],y=b[i];
  if(x>y)
  swap(x,y);
  if(lower_bound(c+1,c+2*n+1,x)-c<=n-2)
  {
   if(lower_bound(c+1,c+2*n+1,y)-c<=n-1)
   ans=min(ans,s[n]);
   else
   ans=min(ans,y+s[n-1]);
  }
  else
  ans=min(ans,x+y+s[n-2]);
 }
 printf("%lld\n",ans);
 return 0;
}
```