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>
#include<utility>

using namespace std;

int main(){
	int n;
	cin >> n;
	vector<pair<int, int>> p;
	long long as = 0, bs = 0, ans = 0;
	for(int i = 0; i < n; i++){
		int a, b;
		cin >> a >> b;
		p.push_back({a, i});
		p.push_back({b, i});
		as += a;
		bs += b;
	}
	ans = min(as, bs);
	sort(p.begin(), p.end());
	vector<int> e(n);
	for(int i = 0; i < n; i++){
		e[p[i].second] = 1;
	}
	if(find(e.begin(), e.end(), 0) == e.end()){
		int q1, q2, q3, q4, d1, d2;
		if(p[n - 1].second != p[n].second){
			q1 = n - 1, q2 = n;
		}else{
			q1 = n - 1, q2 = n + 1;
		}
		if(p[n - 2].second != p[n].second){
			q3 = n - 2, q4 = n;
		} else{
			q3 = n - 2, q4 = n + 1;
		}
		d1 = p[q2].first - p[q1].first;
		d2 = p[q4].first - p[q3].first;
		if(d1 <= d2) swap(p[q1], p[q2]);
		else swap(p[q3], p[q4]);
	}
	long long sum = 0;
	for(int i = 0; i < n; i++){
		sum += p[i].first;
	}
	ans = min(ans, sum);
	cout << ans << endl;
	return 0;
}