#include <bits/stdc++.h>
#define for1(a,b,i) for(int i=a;i<=b;++i)
#define FOR2(a,b,i) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
inline int read() {
    int f=1,sum=0;
    char x=getchar();
    for(;(x<'0'||x>'9');x=getchar()) if(x=='-') f=-1;
    for(;x>='0'&&x<='9';x=getchar()) sum=sum*10+x-'0';
    return f*sum;
}
int n;
int a[500005][2],sta[500005];
int main(){
    n=read();
    for1(1,n,i) {
        a[i][0]=read();
        a[i][1]=read();
        sta[i*2]=a[i][1];
        sta[i*2-1]=a[i][0];
    }
    sort(sta+1,sta+2*n+1);
    
    ll ans=0,sum=0;
    for1(1,n,i) ans+=a[i][0],sum+=a[i][1];
    ans=min(ans,sum),sum=0;
    for1(1,n,i) sum+=sta[i];
    for1(1,n,i) {
        if(max(a[i][0],a[i][1])<sta[n]) return printf("%lld\n",sum),0;
        if(min(a[i][0],a[i][1])>sta[n]) return printf("%lld\n",sum),0;
        ll now=sum;
        now+=max(a[i][0],a[i][1])-sta[n];
        now+=max(0,min(a[i][0],a[i][1])-sta[n-1]);
        ans=min(ans,now);
    }
    printf("%lld\n",ans);
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
#define for1(a,b,i) for(int i=a;i<=b;++i)
#define FOR2(a,b,i) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
inline int read() {
    int f=1,sum=0;
    char x=getchar();
    for(;(x<'0'||x>'9');x=getchar()) if(x=='-') f=-1;
    for(;x>='0'&&x<='9';x=getchar()) sum=sum*10+x-'0';
    return f*sum;
}
int n;
int a[500005][2],sta[500005];
int main(){
    n=read();
    for1(1,n,i) {
        a[i][0]=read();
        a[i][1]=read();
        sta[i*2]=a[i][1];
        sta[i*2-1]=a[i][0];
    }
    sort(sta+1,sta+2*n+1);
    
    ll ans=0,sum=0;
    for1(1,n,i) ans+=a[i][0],sum+=a[i][1];
    ans=min(ans,sum),sum=0;
    for1(1,n,i) sum+=sta[i];
    for1(1,n,i) {
        if(max(a[i][0],a[i][1])<sta[n]) return printf("%lld\n",sum),0;
        if(min(a[i][0],a[i][1])>sta[n]) return printf("%lld\n",sum),0;
        ll now=sum;
        now+=max(a[i][0],a[i][1])-sta[n];
        now+=max(0,min(a[i][0],a[i][1])-sta[n-1]);
        ans=min(ans,now);
    }
    printf("%lld\n",ans);
}
```