#include<bits/stdc++.h>
#define ll long long
#define db double
using namespace std;
 
const int inf = 1e9;
const ll INF = 1e18;

const int mn = 1e5 + 10;
const ll mod = 1e9 + 7;

ll a[mn], b[mn], c[2 * mn];
ll pre[2 * mn];
int main()
{
	int n;	scanf("%d", &n);
	ll tans1 = 0, tans2 = 0;
	for (int i = 1; i <= n; i++)
	{
		scanf("%lld %lld", &a[i], &b[i]);
		tans1 += a[i], tans2 += b[i];
		c[i] = a[i], c[i + n] = b[i];
	}
	
	ll ans = min(tans1, tans2);
	
	sort(c + 1, c + 2 * n + 1);
	for (int i = 1; i <= n; i++)
		pre[i] = pre[i - 1] + c[i];
	for (int i = 1; i <= n; i++)
	{
		ll x = a[i], y = b[i];
		if (x > y)	swap(x, y);
		
		int p1 = lower_bound(c + 1, c + 2 * n + 1, x) - c;
		int p2 = lower_bound(c + 1, c + 2 * n + 1, y) - c;
		
		if (p1 > n - 2 && p2 > n - 2)
			ans = min(ans, x + y + pre[n - 2]);
		else if (p1 <= n - 1 && p2 > n - 1)
			ans = min(ans, pre[n - 1] + y);
		else if (p1 <= n && p2 <= n)
			ans = min(ans, pre[n]);
	}
	printf("%lld\n", ans);
	
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 ll long long
#define db double
using namespace std;
 
const int inf = 1e9;
const ll INF = 1e18;

const int mn = 1e5 + 10;
const ll mod = 1e9 + 7;

ll a[mn], b[mn], c[2 * mn];
ll pre[2 * mn];
int main()
{
	int n;	scanf("%d", &n);
	ll tans1 = 0, tans2 = 0;
	for (int i = 1; i <= n; i++)
	{
		scanf("%lld %lld", &a[i], &b[i]);
		tans1 += a[i], tans2 += b[i];
		c[i] = a[i], c[i + n] = b[i];
	}
	
	ll ans = min(tans1, tans2);
	
	sort(c + 1, c + 2 * n + 1);
	for (int i = 1; i <= n; i++)
		pre[i] = pre[i - 1] + c[i];
	for (int i = 1; i <= n; i++)
	{
		ll x = a[i], y = b[i];
		if (x > y)	swap(x, y);
		
		int p1 = lower_bound(c + 1, c + 2 * n + 1, x) - c;
		int p2 = lower_bound(c + 1, c + 2 * n + 1, y) - c;
		
		if (p1 > n - 2 && p2 > n - 2)
			ans = min(ans, x + y + pre[n - 2]);
		else if (p1 <= n - 1 && p2 > n - 1)
			ans = min(ans, pre[n - 1] + y);
		else if (p1 <= n && p2 <= n)
			ans = min(ans, pre[n]);
	}
	printf("%lld\n", ans);
	
	return 0;
}

```