#include<bits/stdc++.h>
#define MAX 100005
using namespace std;
 
int main(){
  int n;
  long long ans = 0, ans2, ans3, a[MAX*2], c, d;
  scanf("%d%lld%lld", &n, &c, &d);
  ans2 = c;
  ans3 = d;
  for(int i = 0; i < n-1; i++){
    scanf("%lld%lld", &a[i*2], &a[i*2+1]);
    ans2 += a[i*2];
    ans3 += a[i*2+1];
    if(max(a[i*2], a[i*2+1]) < max(c, d)){
      swap(a[i*2], c);
      swap(a[i*2+1], d);
    }
  }
  sort(a, a+2*n-2);
  for(int i = 0; i < n-2; i++) ans += a[i];
  ans += min(c + d, a[n-2] + a[n-1]);
  printf("%lld\n", min(ans, min(ans2, ans3)));
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 MAX 100005
using namespace std;
 
int main(){
  int n;
  long long ans = 0, ans2, ans3, a[MAX*2], c, d;
  scanf("%d%lld%lld", &n, &c, &d);
  ans2 = c;
  ans3 = d;
  for(int i = 0; i < n-1; i++){
    scanf("%lld%lld", &a[i*2], &a[i*2+1]);
    ans2 += a[i*2];
    ans3 += a[i*2+1];
    if(max(a[i*2], a[i*2+1]) < max(c, d)){
      swap(a[i*2], c);
      swap(a[i*2+1], d);
    }
  }
  sort(a, a+2*n-2);
  for(int i = 0; i < n-2; i++) ans += a[i];
  ans += min(c + d, a[n-2] + a[n-1]);
  printf("%lld\n", min(ans, min(ans2, ans3)));
}

```