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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  long long sa = 0;
  long long sb = 0;
  vector<pair<int, int>> c(2 * n);
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    sa += x;
    sb += y;
    c[2 * i] = {x, i};
    c[2 * i + 1] = {y, i};
  }
  sort(c.begin(), c.end());
  long long ret = 0;
  vector<bool> ckd(n, false);
  for (int i = 0; i < n; i++) {
    ret += c[i].first;
    ckd[c[i].second] = true;
  }
  bool flag = true;
  for (int i = 0; i < n; i++) {
    if (!ckd[i]) {
      flag = false;
    }
  }
  if (flag) {
    if (c[n - 1].second != c[n].second) {
      ret = ret - c[n - 1].first + c[n].first;
    } else {
      ret = ret - max(c[n - 1].first - c[n + 1].first, c[n - 2].first - c[n].first);
    }
  }
  ret = min(ret, sa);
  ret = min(ret, sb);
  cout << ret << '\n';
  return 0;
}