#include <bits/stdc++.h>

typedef long long ll;
typedef std::pair <int, int> pr;
const int N = 200054;

int n;
pr a[N];
bool fill[N];

inline ll min(const ll x, const ll y) {return x < y ? x : y;}

int main() {
	int i, u, v; ll ans, s = 0; bool ok = false;
	scanf("%d", &n);
	for (i = 0; i < n; ++i) scanf("%d%d", &u, &v), a[i] = pr(u, i), a[i + n] = pr(v, i + n);
	std::sort(a, a + 2 * n);
	for (u = i = 0; i < n; ++i) s += a[i].first, v = a[i].second % n, fill[v] ? ok = true : fill[v] = true, u += a[i].second < n;
	if (ok |= !u || u == n) ans = 0;
	else if (a[n].second % n == a[n - 1].second % n) ans = min(a[n].first - a[n - 2].first, a[n + 1].first - a[n - 1].first);
	else ans = a[n].first - a[n - 1].first;
	printf("%lld\n", ans + s);
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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;
typedef std::pair <int, int> pr;
const int N = 200054;

int n;
pr a[N];
bool fill[N];

inline ll min(const ll x, const ll y) {return x < y ? x : y;}

int main() {
	int i, u, v; ll ans, s = 0; bool ok = false;
	scanf("%d", &n);
	for (i = 0; i < n; ++i) scanf("%d%d", &u, &v), a[i] = pr(u, i), a[i + n] = pr(v, i + n);
	std::sort(a, a + 2 * n);
	for (u = i = 0; i < n; ++i) s += a[i].first, v = a[i].second % n, fill[v] ? ok = true : fill[v] = true, u += a[i].second < n;
	if (ok |= !u || u == n) ans = 0;
	else if (a[n].second % n == a[n - 1].second % n) ans = min(a[n].first - a[n - 2].first, a[n + 1].first - a[n - 1].first);
	else ans = a[n].first - a[n - 1].first;
	printf("%lld\n", ans + s);
	return 0;
}
```