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

const int N = 2e5 + 5;

struct NODE {
  int val, who, id;
  friend bool operator < (NODE a, NODE b) {
    return a.val < b.val;
  }
}P[N];

int n, m, x, y, A[N][2], tot;

bool vis[N][2];

int main() {
  cin >> n;
  for(int i = 1; i <= n; ++ i) {
    scanf("%d%d", &A[i][0], &A[i][1]);
    P[++ tot] = (NODE){A[i][0], i, 0};
    P[++ tot] = (NODE){A[i][1], i, 1};
  }
  sort(P + 1, P + tot + 1);
  long long ans = 0;
  for(int i = 1; i <= n; ++ i) {
    vis[P[i].who][P[i].id] = 1;
    ans += P[i].val;
  }
  int l = 0, r = 0;
  for(int i = 1; i <= n; ++ i) {
    l += vis[i][0];
    r += vis[i][1];
    if(vis[i][0] == vis[i][1]) {
      printf("%lld\n", ans);
      return 0;
    }
  }
  if(l == n || r == n) {
    cout << ans << endl;
    return 0;
  }

  int res = 1e9;

  vis[P[n + 1].who][P[n + 1].id] = 1;
  for(int i = 1; i <= n; ++ i) {
    if(!vis[P[i].who][P[i].id ^ 1]) {
      res = min(res, P[n + 1].val - P[i].val);
    }
    else res = min(res, P[n + 2].val - P[i].val);
  }
  cout << ans + res << endl;
}
