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

typedef long long ll;
int n, c[500233];
pair<int, int> a[1000233];

int main() {
  scanf("%d", &n);
  ll s1 = 0, s2 = 0;
  for (int i = 1, x, y; i <= n; i++) {
    scanf("%d%d", &x, &y);
    a[i] = make_pair(x, i);
    a[i + n] = make_pair(y, i);
    s1 += x;
    s2 += y;
  }
  sort(a + 1, a + 1 + n * 2, greater<pair<int, int>>());
  ll ans = max(s1, s2), cur = 0;
  int ok = 0;
  for (int i = 1; i <= n; i++) {
    if (c[a[i].second]) ok = 1;
    c[a[i].second] = 1;
    cur += a[i].first;
  }
  if (ok) {
    ans = max(ans, cur);
  } else {
    for (int i = n + 1; i <= n * 2; i++) {
      if (a[i].second != a[n].second) {
        ans = max(ans, cur + a[i].first - a[n].first);
      } else {
        ans = max(ans, cur + a[i].first - a[n - 1].first);
      }
    }
  }
  printf("%lld\n", s1 + s2 - ans);
}

### 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;

typedef long long ll;
int n, c[500233];
pair<int, int> a[1000233];

int main() {
  scanf("%d", &n);
  ll s1 = 0, s2 = 0;
  for (int i = 1, x, y; i <= n; i++) {
    scanf("%d%d", &x, &y);
    a[i] = make_pair(x, i);
    a[i + n] = make_pair(y, i);
    s1 += x;
    s2 += y;
  }
  sort(a + 1, a + 1 + n * 2, greater<pair<int, int>>());
  ll ans = max(s1, s2), cur = 0;
  int ok = 0;
  for (int i = 1; i <= n; i++) {
    if (c[a[i].second]) ok = 1;
    c[a[i].second] = 1;
    cur += a[i].first;
  }
  if (ok) {
    ans = max(ans, cur);
  } else {
    for (int i = n + 1; i <= n * 2; i++) {
      if (a[i].second != a[n].second) {
        ans = max(ans, cur + a[i].first - a[n].first);
      } else {
        ans = max(ans, cur + a[i].first - a[n - 1].first);
      }
    }
  }
  printf("%lld\n", s1 + s2 - ans);
}

```