#include <bits/stdc++.h>

typedef long long LL;
const int N = 200005;

int n;
bool vis[N];
LL A, B, ans;
std::pair<int, int> a[N];

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n;
	for (int i = 0, x, y; i < n; ++i)
		std::cin >> x >> y, a[i] = {x, i}, a[i + n] = {y, i}, A += x, B += y;
	ans = std::min(A, B), std::sort(a, a + n + n);
	for (int i = 0; i < n; ++i) vis[a[i].second] = 1;
	LL sum = 0;
	for (int i = 0; i < n; ++i) sum += a[i].first;
	for (int i = 0; i < n; ++i) if (!vis[i])
		return std::cout << std::min(ans, sum) << '\n', 0;
	if (a[n - 1].second != a[n].second)
		std::cout << std::min(ans, sum - a[n - 1].first + a[n].first) << '\n';
	else
		std::cout << std::min(ans, sum + std::min(a[n].first - a[n - 2].first, a[n + 1].first - a[n - 1].first)) << '\n';
	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>

typedef long long LL;
const int N = 200005;

int n;
bool vis[N];
LL A, B, ans;
std::pair<int, int> a[N];

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n;
	for (int i = 0, x, y; i < n; ++i)
		std::cin >> x >> y, a[i] = {x, i}, a[i + n] = {y, i}, A += x, B += y;
	ans = std::min(A, B), std::sort(a, a + n + n);
	for (int i = 0; i < n; ++i) vis[a[i].second] = 1;
	LL sum = 0;
	for (int i = 0; i < n; ++i) sum += a[i].first;
	for (int i = 0; i < n; ++i) if (!vis[i])
		return std::cout << std::min(ans, sum) << '\n', 0;
	if (a[n - 1].second != a[n].second)
		std::cout << std::min(ans, sum - a[n - 1].first + a[n].first) << '\n';
	else
		std::cout << std::min(ans, sum + std::min(a[n].first - a[n - 2].first, a[n + 1].first - a[n - 1].first)) << '\n';
	return 0;
}
```