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 n;
pair<int,int> a[200005];
int had[100005];

int id(int i) { return abs(a[i].second); }

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d %d", &a[i].first, &a[i + n].first);
		a[i].second = i;
		a[i + n].second = -i;
	}
	sort(a + 1, a + n + n + 1);
	long long res = 0;
	for (int i = 1; i <= n; ++i) res += a[i].first;
	int had_a = 0, had_b = 0;
	for (int i = 1; i <= n; ++i) {
		had[id(i)] = 1;
		if (a[i].second < 0) had_b++;
		else had_a++;
	}
	int nhad = 0;
	for (int i = 1; i <= n; ++i) nhad += had[i];
	if (had_a == n || had_b == n || nhad < n) return cout << res << '\n',0;
	if (id(n) != id(n + 1) 
	|| (had_a - (a[n].second > 0) + (a[n + 1].second > 0) == 0)
	|| (had_b - (a[n].second < 0) + (a[n + 1].second < 0) == 0) 
	) return cout << res + a[n + 1].first - a[n].first << '\n',0;
	cout << res + min(a[n + 2].first - a[n].first, a[n + 1].first - a[n - 1].first) << '\n';
	
}
