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;
#define ll long long
#define ii pair<ll,ll>
#define iii pair<ll,ii>
#define endl '\n'

int n;
vector<ii> arr,brr,crr;
bool used[100005];

int main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	
	cin>>n;
	ll a,b;
	for (int x=0;x<n;x++){
		cin>>a>>b;
		arr.push_back(ii(a,x));
		brr.push_back(ii(b,x));
		crr.push_back(ii(a,x));
		crr.push_back(ii(b,x));
	}
	
	//cout<<ans<<endl;
	
	sort(arr.begin(),arr.end());
	sort(brr.begin(),brr.end());
	sort(crr.begin(),crr.end());
	
	bool can=false;
	ll ans=0;
	for (int x=0;x<n;x++){
		ans+=crr[x].first;
		can=can||used[crr[x].second];
		used[crr[x].second]=true;
	}
	
	if (arr.back().first<=brr[0].first || brr.back().first<=arr[0].first) can=true;
	
	if (!can){
		if (crr[n-1].second!=crr[n].second) ans+=crr[n].first-crr[n-1].first;
		else ans+=min(crr[n+1].first-crr[n-1].first,crr[n].first-crr[n-2].first);
	}
	
	cout<<ans<<endl;
}