#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 100005;

int a[MAXN], b[MAXN];

int n;

vector<int> v;

ll sum[MAXN], sum1 = 0, sum2 = 0, res;

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d%d", &a[i], &b[i]);
		sum1 += a[i];
		sum2 += b[i];
		v.emplace_back(a[i]);
		v.emplace_back(b[i]);
	}
	res = min(sum1, sum2);
	sort(v.begin(), v.end());
	for (int i = 1; i <= n; i++)
		sum[i] = sum[i - 1] + v[i - 1];
	for (int i = 1; i <= n; i++)
	{
		int x = a[i], y = b[i];
		if (x > y) swap(x, y);
		if (x >= v[n - 1])
			res = min(res, sum[n - 2] + x + y);
		else if (y > v[n - 1])
			res = min(res, sum[n - 1] + y);
		else
			res = min(res, sum[n]);
	}
	printf("%lld\n", res);
	return 0;
}

### Prompt

Please formulate 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;

typedef long long ll;

const int MAXN = 100005;

int a[MAXN], b[MAXN];

int n;

vector<int> v;

ll sum[MAXN], sum1 = 0, sum2 = 0, res;

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d%d", &a[i], &b[i]);
		sum1 += a[i];
		sum2 += b[i];
		v.emplace_back(a[i]);
		v.emplace_back(b[i]);
	}
	res = min(sum1, sum2);
	sort(v.begin(), v.end());
	for (int i = 1; i <= n; i++)
		sum[i] = sum[i - 1] + v[i - 1];
	for (int i = 1; i <= n; i++)
	{
		int x = a[i], y = b[i];
		if (x > y) swap(x, y);
		if (x >= v[n - 1])
			res = min(res, sum[n - 2] + x + y);
		else if (y > v[n - 1])
			res = min(res, sum[n - 1] + y);
		else
			res = min(res, sum[n]);
	}
	printf("%lld\n", res);
	return 0;
}

```