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

int n, vis[101010];
vector<pair<int, int> > v;
long long mx, my, Ans;

int main(void) {
  scanf("%d", &n);
  for(int i = 1, x, y; i <= n; i++) {
    scanf("%d%d", &x, &y);
    mx += x;
    my += y;
    v.push_back({x, i});
    v.push_back({y, i});
  }
  
  sort(v.begin(), v.end());
  bool flg = 0;
  for(int i = 0; i < n; i++) {
    Ans += v[i].first;
    if(vis[v[i].second]) flg = 1;
    else vis[v[i].second] = 1;
  }
  if(!flg) {
    if(v[n].second == v[n - 1].second) 
      Ans = min(Ans - v[n - 1].first + v[n + 1].first,
		Ans - v[n - 2].first + v[n].first);
    else Ans = Ans - v[n - 1].first + v[n].first;
  }
  cout << min(Ans, min(mx, my)) << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;

int n, vis[101010];
vector<pair<int, int> > v;
long long mx, my, Ans;

int main(void) {
  scanf("%d", &n);
  for(int i = 1, x, y; i <= n; i++) {
    scanf("%d%d", &x, &y);
    mx += x;
    my += y;
    v.push_back({x, i});
    v.push_back({y, i});
  }
  
  sort(v.begin(), v.end());
  bool flg = 0;
  for(int i = 0; i < n; i++) {
    Ans += v[i].first;
    if(vis[v[i].second]) flg = 1;
    else vis[v[i].second] = 1;
  }
  if(!flg) {
    if(v[n].second == v[n - 1].second) 
      Ans = min(Ans - v[n - 1].first + v[n + 1].first,
		Ans - v[n - 2].first + v[n].first);
    else Ans = Ans - v[n - 1].first + v[n].first;
  }
  cout << min(Ans, min(mx, my)) << endl;
  return 0;
}

```