#include <bits/stdc++.h>
#define il inline
#define ll long long
const int N=2e5+5;

int n,s[2],o,v[N]; ll ans; struct node{int v,id,o;}d[N];

il bool cmp(node x,node y){return x.v<y.v;}

int main()
{
	scanf("%d",&n); int i;
	for (i=1; i<=n; i++)
		scanf("%d%d",&d[i+i-1].v,&d[i+i].v),
		d[i+i-1].id=d[i+i].id=i,d[i+i].o=1;
	std::sort(d+1,d+n+n+1,cmp);
	
	for (i=1; i<=n; i++) ans+=(ll)d[i].v,s[d[i].o]++,o|=++v[d[i].id]>1;
	if (o||s[0]==n||s[1]==n) return printf("%lld",ans),0;
	ans+=(ll)d[n+1].v-d[n].v,s[d[n].o]--,s[d[n+1].o]++;
	if (d[n].id!=d[n+1].id||s[0]==n||s[1]==n) return printf("%lld",ans),0;
	printf("%lld",ans+(ll)std::min(d[n+2].v-d[n+1].v,d[n].v-d[n-1].v));
	
	return 0;
}
### Prompt

Create a solution in Cpp for the following problem:

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 il inline
#define ll long long
const int N=2e5+5;

int n,s[2],o,v[N]; ll ans; struct node{int v,id,o;}d[N];

il bool cmp(node x,node y){return x.v<y.v;}

int main()
{
	scanf("%d",&n); int i;
	for (i=1; i<=n; i++)
		scanf("%d%d",&d[i+i-1].v,&d[i+i].v),
		d[i+i-1].id=d[i+i].id=i,d[i+i].o=1;
	std::sort(d+1,d+n+n+1,cmp);
	
	for (i=1; i<=n; i++) ans+=(ll)d[i].v,s[d[i].o]++,o|=++v[d[i].id]>1;
	if (o||s[0]==n||s[1]==n) return printf("%lld",ans),0;
	ans+=(ll)d[n+1].v-d[n].v,s[d[n].o]--,s[d[n+1].o]++;
	if (d[n].id!=d[n+1].id||s[0]==n||s[1]==n) return printf("%lld",ans),0;
	printf("%lld",ans+(ll)std::min(d[n+2].v-d[n+1].v,d[n].v-d[n-1].v));
	
	return 0;
}
```