#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 1e5 + 10;

typedef pair<int, int> pi;

int n, a[N], b[N], num[N];
vector<pi> p;
ll ans, cur;

int main() {
  scanf("%d", &n);
  ll suma = 0, sumb = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &a[i], &b[i]);
    p.push_back(pi(a[i], i));
    p.push_back(pi(b[i], i));
    suma += a[i];
    sumb += b[i];
  }
  sort(p.begin(), p.end());
  for (int i = 0; i < n; i++) {
    cur += p[i].first;
    num[p[i].second]++;
  }
  for (int i = 1; i <= n; i++) if (num[i] == 2) {
    cout << cur << endl;
    return 0;
  }
  ans = min(suma, sumb);
  for (int i = 0; i < n; i++) {
    if (i != n - 1) {
      ans = min(ans, cur - p[i].first + a[p[i].second] + b[p[i].second] - p[n - 1].first);
    } else {
      ans = min(ans, cur - p[i].first + a[p[i].second] + b[p[i].second] - p[n - 2].first);
    }
  }
  cout << ans << endl;
  return 0;
}

### 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>

using namespace std;

typedef long long ll;

const int N = 1e5 + 10;

typedef pair<int, int> pi;

int n, a[N], b[N], num[N];
vector<pi> p;
ll ans, cur;

int main() {
  scanf("%d", &n);
  ll suma = 0, sumb = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &a[i], &b[i]);
    p.push_back(pi(a[i], i));
    p.push_back(pi(b[i], i));
    suma += a[i];
    sumb += b[i];
  }
  sort(p.begin(), p.end());
  for (int i = 0; i < n; i++) {
    cur += p[i].first;
    num[p[i].second]++;
  }
  for (int i = 1; i <= n; i++) if (num[i] == 2) {
    cout << cur << endl;
    return 0;
  }
  ans = min(suma, sumb);
  for (int i = 0; i < n; i++) {
    if (i != n - 1) {
      ans = min(ans, cur - p[i].first + a[p[i].second] + b[p[i].second] - p[n - 1].first);
    } else {
      ans = min(ans, cur - p[i].first + a[p[i].second] + b[p[i].second] - p[n - 2].first);
    }
  }
  cout << ans << endl;
  return 0;
}

```