#include <cstdio>

#include <algorithm>
#include <numeric>

using namespace std;

typedef long long ll;

const int N = 100010;

int n;
int a[N], b[N];
int arr[N * 2];
ll sum[N * 2];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf("%d%d", &a[i], &b[i]);
	ll ans = min(accumulate(a + 1, a + n + 1, 0LL), accumulate(b + 1, b + n + 1, 0LL));
	copy(a + 1, a + n + 1, arr + 1);
	copy(b + 1, b + n + 1, arr + n + 1);
	sort(arr + 1, arr + n * 2 + 1);
	for (int i = 1; i <= n * 2; ++i)
		sum[i] = sum[i - 1] + arr[i];
	for (int i = 1; i <= n; ++i) {
		int x = lower_bound(arr + 1, arr + n * 2 + 1, a[i]) - arr,
		    y = lower_bound(arr + 1, arr + n * 2 + 1, b[i]) - arr;
		if (x == y)
			++y;
		for (int j = n; j >= n - 2; --j)
			if ((j + (x > j) + (y > j)) == n)
				ans = min(ans, sum[j] + (x > j) * arr[x] + (y > j) * arr[y]);
	}
	printf("%lld\n", ans);
	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 <cstdio>

#include <algorithm>
#include <numeric>

using namespace std;

typedef long long ll;

const int N = 100010;

int n;
int a[N], b[N];
int arr[N * 2];
ll sum[N * 2];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf("%d%d", &a[i], &b[i]);
	ll ans = min(accumulate(a + 1, a + n + 1, 0LL), accumulate(b + 1, b + n + 1, 0LL));
	copy(a + 1, a + n + 1, arr + 1);
	copy(b + 1, b + n + 1, arr + n + 1);
	sort(arr + 1, arr + n * 2 + 1);
	for (int i = 1; i <= n * 2; ++i)
		sum[i] = sum[i - 1] + arr[i];
	for (int i = 1; i <= n; ++i) {
		int x = lower_bound(arr + 1, arr + n * 2 + 1, a[i]) - arr,
		    y = lower_bound(arr + 1, arr + n * 2 + 1, b[i]) - arr;
		if (x == y)
			++y;
		for (int j = n; j >= n - 2; --j)
			if ((j + (x > j) + (y > j)) == n)
				ans = min(ans, sum[j] + (x > j) * arr[x] + (y > j) * arr[y]);
	}
	printf("%lld\n", ans);
	return 0;
}

```