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<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;

using namespace std;
#define N 100010
ll n,a[N],b[N];
int main(){
    cin>>n;
    vector<pair<ll,ll> > v;
    lol(i,n){
	cin>>a[i]>>b[i];
	v.push_back(make_pair(a[i],i*2));
	v.push_back(make_pair(b[i],i*2+1));
    }
    sort(v.begin(),v.end());
    ll k=v[n-1].first;
    bool ok=1,can=0;
    lol(i,n)ok&=(a[i]<=k);
    can|=ok;ok=1;
    lol(i,n)ok&=(b[i]<=k);
    can|=ok;ok=0;
    lol(i,n)ok|=(a[i]<=k&&b[i]<=k);
    can|=ok;
    ll sum=0;
    lol(i,n)sum+=v[i].first;
    if(can){
	cout<<sum<<endl;
	return 0;
    }
    if(v[n-1].second/2!=v[n].second/2){
	cout<<sum-v[n-1].first+v[n].first<<endl;
	return 0;
    }
    ll vl=sum-v[n-1].first+v[n+1].first;
    ll vr=sum-v[n-2].first+v[n].first;
    cout<<min(vl,vr)<<endl;
    return 0;
}
