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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<ll, P> Q;
vector<Q> vec;
int n, p[100010], num;
ll a, b, ans;
bool jud;

int main()
{
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		cin >> a >> b;
		vec.push_back(Q(a, P(i, 1)));
		vec.push_back(Q(b, P(i, 2)));
	}
	sort(vec.begin(), vec.end());
	for (int i = 0; i < n; i++)
	{
		p[vec[i].second.first]++;
		if (p[vec[i].second.first] == 2)
		{
			jud = true;
		}
		if (vec[i].second.second == 1)
		{
			num++;
		}
		ans += vec[i].first;
	}
	if (jud || num == n || num == 0)
	{
		cout << ans << endl;
	}
	else if (vec[n - 1].second.first != vec[n].second.first)
	{
		cout << ans - vec[n - 1].first + vec[n].first << endl;
	}
	else if ((num == n - 1 && vec[n].second.second == 1) || (num == 1 && vec[n].second.second == 2))
	{
		cout << ans - vec[n - 1].first + vec[n].first << endl;
	}
	else
	{
		cout << ans + min(vec[n].first - vec[n - 2].first, vec[n + 1].first - vec[n - 1].first) << endl;
	}
	cin >> n;
}