#include <bits/stdc++.h>

using namespace std;

const int MX = 100000;

int a[MX], b[MX];
long long s[2 * MX];

int main() {
	int n;
	ignore = scanf("%d", &n);
	
	long long sa = 0, sb = 0;
	vector<tuple<int, int, int>> vec;
	for (int i = 0; i < n; i++) {
		ignore = scanf("%d %d", a + i, b + i);
		sa += a[i];
		sb += b[i];
		vec.emplace_back(a[i], i, 0);
		vec.emplace_back(b[i], i, 1);
	}
	
	sort(vec.begin(), vec.end());
	s[0] = get<0>(vec[0]);
	for (int i = 1; i < 2 * n; i++) {
		s[i] = s[i - 1] + get<0>(vec[i]);
	}
	
	long long ans = min(sa, sb);
	for (int i = 0; i < n; i++) {
		int x = lower_bound(vec.begin(), vec.end(), make_tuple(a[i], i, 0)) - vec.begin();
		int y = lower_bound(vec.begin(), vec.end(), make_tuple(b[i], i, 1)) - vec.begin();
		int k = n - 2;
		while (k + (x < k ? 0 : 1) + (y < k ? 0 : 1) < n) k++;
		
		long long cur = s[k - 1];
		if (x >= k) cur += a[i];
		if (y >= k) cur += b[i];
		ans = min(ans, cur);
	}
	
	printf("%lld\n", ans);
	
	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>

using namespace std;

const int MX = 100000;

int a[MX], b[MX];
long long s[2 * MX];

int main() {
	int n;
	ignore = scanf("%d", &n);
	
	long long sa = 0, sb = 0;
	vector<tuple<int, int, int>> vec;
	for (int i = 0; i < n; i++) {
		ignore = scanf("%d %d", a + i, b + i);
		sa += a[i];
		sb += b[i];
		vec.emplace_back(a[i], i, 0);
		vec.emplace_back(b[i], i, 1);
	}
	
	sort(vec.begin(), vec.end());
	s[0] = get<0>(vec[0]);
	for (int i = 1; i < 2 * n; i++) {
		s[i] = s[i - 1] + get<0>(vec[i]);
	}
	
	long long ans = min(sa, sb);
	for (int i = 0; i < n; i++) {
		int x = lower_bound(vec.begin(), vec.end(), make_tuple(a[i], i, 0)) - vec.begin();
		int y = lower_bound(vec.begin(), vec.end(), make_tuple(b[i], i, 1)) - vec.begin();
		int k = n - 2;
		while (k + (x < k ? 0 : 1) + (y < k ? 0 : 1) < n) k++;
		
		long long cur = s[k - 1];
		if (x >= k) cur += a[i];
		if (y >= k) cur += b[i];
		ans = min(ans, cur);
	}
	
	printf("%lld\n", ans);
	
	return 0;
}

```