#include<algorithm>
#include<cstdio>
#define N 200000
using namespace std;
int a[N],b[N],c[N],i,n;long long x;
inline bool cmp(int u,int v){return a[u]<a[v];}
int main()
{
    for(scanf("%d",&n),i=0;i<n;i++)scanf("%d%d",a+(b[i<<1]=i<<1),a+(b[i<<1|1]=i<<1|1));
    for(sort(b,b+(n<<1),cmp),i=0;i<n;i++)x+=a[b[i]],c[b[i]>>1]++;
    for(i=0;i<n;i++)if(c[i]==2)return 0&printf("%lld\n",x);
    for(i=1;i<n;i++)if((b[i-1]^b[i])&1)return 0&printf("%lld\n",x+(b[n-1]>>1==b[n]>>1?a[b[n]]-a[b[n-2]]<a[b[n+1]]-a[b[n-1]]?a[b[n]]-a[b[n-2]]:a[b[n+1]]-a[b[n-1]]:a[b[n]]-a[b[n-1]]));
    return 0&printf("%lld\n",x);
}
### Prompt

Please formulate 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<algorithm>
#include<cstdio>
#define N 200000
using namespace std;
int a[N],b[N],c[N],i,n;long long x;
inline bool cmp(int u,int v){return a[u]<a[v];}
int main()
{
    for(scanf("%d",&n),i=0;i<n;i++)scanf("%d%d",a+(b[i<<1]=i<<1),a+(b[i<<1|1]=i<<1|1));
    for(sort(b,b+(n<<1),cmp),i=0;i<n;i++)x+=a[b[i]],c[b[i]>>1]++;
    for(i=0;i<n;i++)if(c[i]==2)return 0&printf("%lld\n",x);
    for(i=1;i<n;i++)if((b[i-1]^b[i])&1)return 0&printf("%lld\n",x+(b[n-1]>>1==b[n]>>1?a[b[n]]-a[b[n-2]]<a[b[n+1]]-a[b[n-1]]?a[b[n]]-a[b[n-2]]:a[b[n+1]]-a[b[n-1]]:a[b[n]]-a[b[n-1]]));
    return 0&printf("%lld\n",x);
}
```