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

const int N = 100005;

int n;
int a[N], b[N];

int used[N];

int main() {
  scanf("%d", &n);
  
  long long sumA = 0, sumB = 0;
  for (int i=0; i<n; i++) {
    scanf("%d %d", a + i, b + i);
    
    sumA += a[i];
    sumB += b[i];
  }
  
  long long ans = min(sumA, sumB);
  
  vector<pair<long long, int>> v;
  for (int i=0; i<n; i++) {
    v.push_back({a[i], i});
    v.push_back({b[i], i});
  }
  sort(v.begin(), v.end());
  
  long long mini = 0;
  for (int i=0; i<n; i++) {
    ++used[v[i].second];
    mini += v[i].first;
  }
  
  for (int i=0; i<n; i++) {
    if (used[i] == 0) ans = min(ans, mini);
    else if (used[i] == 1) {
      if (v[n].second == i) ans = min(ans, mini - min(a[i], b[i]) + v[n+1].first);
      else ans = min(ans, mini - min(a[i], b[i]) + v[n].first);
    } else {
      ans = min(ans, mini - a[i] - b[i] + v[n].first + v[n+1].first);
    }
  }
  
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;

const int N = 100005;

int n;
int a[N], b[N];

int used[N];

int main() {
  scanf("%d", &n);
  
  long long sumA = 0, sumB = 0;
  for (int i=0; i<n; i++) {
    scanf("%d %d", a + i, b + i);
    
    sumA += a[i];
    sumB += b[i];
  }
  
  long long ans = min(sumA, sumB);
  
  vector<pair<long long, int>> v;
  for (int i=0; i<n; i++) {
    v.push_back({a[i], i});
    v.push_back({b[i], i});
  }
  sort(v.begin(), v.end());
  
  long long mini = 0;
  for (int i=0; i<n; i++) {
    ++used[v[i].second];
    mini += v[i].first;
  }
  
  for (int i=0; i<n; i++) {
    if (used[i] == 0) ans = min(ans, mini);
    else if (used[i] == 1) {
      if (v[n].second == i) ans = min(ans, mini - min(a[i], b[i]) + v[n+1].first);
      else ans = min(ans, mini - min(a[i], b[i]) + v[n].first);
    } else {
      ans = min(ans, mini - a[i] - b[i] + v[n].first + v[n+1].first);
    }
  }
  
  printf("%lld\n", ans);
  return 0;
}

```