#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
	int N;
	cin >> N;
	vector<int> A(2 * N), perm(2 * N);
	for (int i = 0; i < N; ++i) cin >> A[i] >> A[i + N];
	for (int i = 0; i < 2 * N; ++i) perm[i] = i;
	sort(perm.begin(), perm.end(), [&](int i, int j) { return A[i] < A[j]; });
	vector<int> c(N);
	long long sum = 0;
	for (int i = 0; i < N; ++i) {
		c[perm[i] % N] |= 1 << (perm[i] / N);
		sum += A[perm[i]];
	}
	vector<int> n(4);
	for (int i = 0; i < N; ++i) ++n[c[i]];
	if (n[1] >= 1 && n[2] >= 1 && n[0] == 0) {
		if (perm[N] % N != perm[N - 1] % N) sum += A[perm[N]] - A[perm[N - 1]];
		else sum += min(A[perm[N + 1]] - A[perm[N - 1]], A[perm[N]] - A[perm[N - 2]]);
	}
	cout << sum << endl;
	return 0;
}
### Prompt

Develop a solution in cpp 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 <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
	int N;
	cin >> N;
	vector<int> A(2 * N), perm(2 * N);
	for (int i = 0; i < N; ++i) cin >> A[i] >> A[i + N];
	for (int i = 0; i < 2 * N; ++i) perm[i] = i;
	sort(perm.begin(), perm.end(), [&](int i, int j) { return A[i] < A[j]; });
	vector<int> c(N);
	long long sum = 0;
	for (int i = 0; i < N; ++i) {
		c[perm[i] % N] |= 1 << (perm[i] / N);
		sum += A[perm[i]];
	}
	vector<int> n(4);
	for (int i = 0; i < N; ++i) ++n[c[i]];
	if (n[1] >= 1 && n[2] >= 1 && n[0] == 0) {
		if (perm[N] % N != perm[N - 1] % N) sum += A[perm[N]] - A[perm[N - 1]];
		else sum += min(A[perm[N + 1]] - A[perm[N - 1]], A[perm[N]] - A[perm[N - 2]]);
	}
	cout << sum << endl;
	return 0;
}
```