#include <bits/stdc++.h>
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const int maxn=(2e5)+10;
int n,d[maxn],a[maxn],b[maxn];
ll ans,A,B,s[maxn];
int main() {
	read(n);
	for (int i=1;i<=n;i++) {
		read(a[i]); read(b[i]);
		A+=a[i]; B+=b[i];
		d[i*2-1]=a[i]; d[i*2]=b[i];
	}
	sort(d+1,d+(2*n+1));
	ans=min(A,B);
	for (int i=1;i<=2*n;i++) s[i]=s[i-1]+d[i];
	for (int i=1;i<=n;i++) {
		int pos=n,tmp=0;
		if (a[i]<=d[pos-1]) pos++; else tmp+=a[i];
		if (b[i]<=d[pos-1]) pos++; else tmp+=b[i];
		ans=min(ans,tmp+s[pos-2]);
	}
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Generate a CPP solution to 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>
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const int maxn=(2e5)+10;
int n,d[maxn],a[maxn],b[maxn];
ll ans,A,B,s[maxn];
int main() {
	read(n);
	for (int i=1;i<=n;i++) {
		read(a[i]); read(b[i]);
		A+=a[i]; B+=b[i];
		d[i*2-1]=a[i]; d[i*2]=b[i];
	}
	sort(d+1,d+(2*n+1));
	ans=min(A,B);
	for (int i=1;i<=2*n;i++) s[i]=s[i-1]+d[i];
	for (int i=1;i<=n;i++) {
		int pos=n,tmp=0;
		if (a[i]<=d[pos-1]) pos++; else tmp+=a[i];
		if (b[i]<=d[pos-1]) pos++; else tmp+=b[i];
		ans=min(ans,tmp+s[pos-2]);
	}
	printf("%lld\n",ans);
	return 0;
}
```