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

struct element
{
   int val, poz, tip;
};

int n;
int A[200005], B[200005], C[200005];
long long S[200005];
long long sa, sb;
int main()
{
   ios_base::sync_with_stdio(false);
   cin.tie(0);
   cin >> n;
   for (int i = 1; i <= n; i++)
   {
      cin >> A[i] >> B[i];
      sa += A[i]; sb += B[i];
      C[2 * i - 1] = A[i];
      C[2 * i] = B[i];
   }

   long long rez = min(sa, sb);

   sort(C + 1, C + 2 * n + 1);

   for (int i = 1; i <= 2 * n; i++)
      S[i] = S[i - 1] + C[i];

   for (int i = 1; i <= n; i++)
   {
      int merg = n;
      long long sum = 0;
      if (A[i] >= C[merg])
         merg--, sum += A[i];
      if (B[i] >= C[merg])
         merg--, sum += B[i];

      rez = min(rez, sum + S[merg]);
   }
   cout << rez;
   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;

struct element
{
   int val, poz, tip;
};

int n;
int A[200005], B[200005], C[200005];
long long S[200005];
long long sa, sb;
int main()
{
   ios_base::sync_with_stdio(false);
   cin.tie(0);
   cin >> n;
   for (int i = 1; i <= n; i++)
   {
      cin >> A[i] >> B[i];
      sa += A[i]; sb += B[i];
      C[2 * i - 1] = A[i];
      C[2 * i] = B[i];
   }

   long long rez = min(sa, sb);

   sort(C + 1, C + 2 * n + 1);

   for (int i = 1; i <= 2 * n; i++)
      S[i] = S[i - 1] + C[i];

   for (int i = 1; i <= n; i++)
   {
      int merg = n;
      long long sum = 0;
      if (A[i] >= C[merg])
         merg--, sum += A[i];
      if (B[i] >= C[merg])
         merg--, sum += B[i];

      rez = min(rez, sum + S[merg]);
   }
   cout << rez;
   return 0;
}

```