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;

const long long INF=10000000000ll;

int main()
{
  int n;
  scanf("%d",&n);
  vector<long long> a(n),b(n);
  for(int i=0;i<n;i++){
    scanf("%lld%lld",&a[i],&b[i]);
  }
  vector<pair<long long,int> > P(2*n);
  for(int i=0;i<n;i++){
    P[i]=make_pair(a[i],i);
    P[i+n]=make_pair(b[i],i+n);
  }
  sort(P.begin(),P.end());
  long long ans=0;
  vector<bool> A(n,false),B(n,false);
  for(int i=0;i<n;i++){
    ans+=P[i].first;
    int k=P[i].second;
    if(k<n)
      A[k]=true;
    else
      B[k-n]=true;
  }
  int L=0,R=0;
  for(int i=0;i<n;i++){
    if(A[i]&&!B[i]){
      L++;
    }
    if(!A[i]&&B[i]){
      R++;
    }
  }
  if(L>0&&R>0&&L+R==n){
    long long S=INF;
    for(int i=0;i<n;i++){
      if(!A[i]){
	S=min(S,a[i]-(L>1&&P[n-1].first==b[i]?P[n-2].first:P[n-1].first));
      }
      if(!B[i]){
	S=min(S,b[i]-(R>1&&P[n-1].first==a[i]?P[n-2].first:P[n-1].first));
      }
    }
    ans+=S;
  }
  printf("%lld\n",ans);
  return 0;
}
