#include<bits/stdc++.h>

using namespace std;

using int64 = long long;
using pi = pair< int, int >;
const int64 INF = 1LL << 60;

int main() {
  int N;
  cin >> N;
  vector< pi > A;
  int64 latte = 0, malta = 0;
  for(int i = 0; i < N; i++) {
    int x, y;
    cin >> x >> y;
    A.emplace_back(x, i);
    A.emplace_back(y, i);
    latte += x;
    malta += y;
  }
  sort(begin(A), end(A));
  bool f = false;
  int used[100000] = {};
  int64 sum = 0;
  for(int i = 0; i < N; i++) {
    if(used[A[i].second]++ >= 1) f = true;
    sum += A[i].first;
  }
  int64 ret = f ? sum : INF;
  for(int i = 0; i < N; i++) {
    for(int j = N; j < N + 2; j++) {
      if(A[i].second == A[j].second) {
        if(i + 1 == N) {
          ret = min(ret, sum - A[N - 2].first + A[j].first);
        } else {
          ret = min(ret, sum - A[N - 1].first + A[j].first);
        }
        break;
      }
    }
  }
  cout << min(ret, min(latte, malta)) << endl;
}

### Prompt

Generate 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<bits/stdc++.h>

using namespace std;

using int64 = long long;
using pi = pair< int, int >;
const int64 INF = 1LL << 60;

int main() {
  int N;
  cin >> N;
  vector< pi > A;
  int64 latte = 0, malta = 0;
  for(int i = 0; i < N; i++) {
    int x, y;
    cin >> x >> y;
    A.emplace_back(x, i);
    A.emplace_back(y, i);
    latte += x;
    malta += y;
  }
  sort(begin(A), end(A));
  bool f = false;
  int used[100000] = {};
  int64 sum = 0;
  for(int i = 0; i < N; i++) {
    if(used[A[i].second]++ >= 1) f = true;
    sum += A[i].first;
  }
  int64 ret = f ? sum : INF;
  for(int i = 0; i < N; i++) {
    for(int j = N; j < N + 2; j++) {
      if(A[i].second == A[j].second) {
        if(i + 1 == N) {
          ret = min(ret, sum - A[N - 2].first + A[j].first);
        } else {
          ret = min(ret, sum - A[N - 1].first + A[j].first);
        }
        break;
      }
    }
  }
  cout << min(ret, min(latte, malta)) << endl;
}

```