#include <bits/stdc++.h>

#ifdef __WIN32
#define LLFORMAT "I64"
#else
#define LLFORMAT "ll"
#endif

using namespace std;

int main() {
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	vector<int> a(n), b(n), c;
	vector<long long> s(n * 2 + 1, 0);
	long long x = 0, y = 0;
	for (int i = 0; i < n; ++i) cin >> a[i] >> b[i], c.push_back(a[i]), c.push_back(b[i]), x += a[i], y += b[i];
	long long ans = min(x, y);
	sort(c.begin(), c.end());
	for (int i = 0; i < n * 2; ++i) s[i + 1] = s[i] + c[i];
	for (int i = 0; i < n; ++i) {
		int x = lower_bound(c.begin(), c.end(), a[i]) - c.begin();
		int y = lower_bound(c.begin(), c.end(), b[i]) - c.begin();
		if(x == y) ++y;
		if(x > y) swap(x, y);
		int t = n - 2;
		long long sum = a[i] + b[i];
		if(x < t) sum -= c[x], ++t;
		if(y < t) sum -= c[y], ++t;
		sum += s[t];
		ans = min(ans, sum);
	}
	cout << ans << endl;
	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>

#ifdef __WIN32
#define LLFORMAT "I64"
#else
#define LLFORMAT "ll"
#endif

using namespace std;

int main() {
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	vector<int> a(n), b(n), c;
	vector<long long> s(n * 2 + 1, 0);
	long long x = 0, y = 0;
	for (int i = 0; i < n; ++i) cin >> a[i] >> b[i], c.push_back(a[i]), c.push_back(b[i]), x += a[i], y += b[i];
	long long ans = min(x, y);
	sort(c.begin(), c.end());
	for (int i = 0; i < n * 2; ++i) s[i + 1] = s[i] + c[i];
	for (int i = 0; i < n; ++i) {
		int x = lower_bound(c.begin(), c.end(), a[i]) - c.begin();
		int y = lower_bound(c.begin(), c.end(), b[i]) - c.begin();
		if(x == y) ++y;
		if(x > y) swap(x, y);
		int t = n - 2;
		long long sum = a[i] + b[i];
		if(x < t) sum -= c[x], ++t;
		if(y < t) sum -= c[y], ++t;
		sum += s[t];
		ans = min(ans, sum);
	}
	cout << ans << endl;
	return 0;
}

```