#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MX = 100005;

pair<int, int> A[2 * MX];

int s[MX];
int cnt[MX];


int main()
{
	int n;
	ignore = scanf("%d", &n);
	
	ll sumA = 0, sumB = 0;
	
	for (int i = 0; i < n; i++) {
		int a, b;
		ignore = scanf("%d %d", &a, &b);
		sumA += a;
		sumB += b;
		A[2 * i] = {a, i};
		A[2 * i + 1] = {b, i};
		s[i] = a + b;
	}

	ll ans = min(sumA, sumB);
		
	sort(A, A + 2 * n);
	
	ll sum = 0;
	int mx1 = 0, mx2 = 0;
	for (int i = 0; i < n; i++) {
		sum += A[i].first;
		cnt[A[i].second]++;
			
		if (A[i].first >= mx1) {
			mx2 = mx1;
			mx1 = A[i].first;
		}
		else if (A[i].first >= mx2) {
			mx2 = A[i].first;
		}
	}
	
	bool found = false;
	for (int i = 0; i < n; i++) {
		if (cnt[i] == 2) {
			found = true;
			break;
		}
	}
	
	if (found) {
		ans = min(ans, sum);
	}
	else {		
		for (int i = 0; i < n; i++) {
			if (A[i].first == mx1) {
				ans = min(ans, sum + s[A[i].second] - A[i].first- mx2);
			}
			else {
				ans = min(ans, sum + s[A[i].second] - A[i].first - mx1);
			}
		}
	}
	
	cout << ans;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
typedef long long ll;

const int MX = 100005;

pair<int, int> A[2 * MX];

int s[MX];
int cnt[MX];


int main()
{
	int n;
	ignore = scanf("%d", &n);
	
	ll sumA = 0, sumB = 0;
	
	for (int i = 0; i < n; i++) {
		int a, b;
		ignore = scanf("%d %d", &a, &b);
		sumA += a;
		sumB += b;
		A[2 * i] = {a, i};
		A[2 * i + 1] = {b, i};
		s[i] = a + b;
	}

	ll ans = min(sumA, sumB);
		
	sort(A, A + 2 * n);
	
	ll sum = 0;
	int mx1 = 0, mx2 = 0;
	for (int i = 0; i < n; i++) {
		sum += A[i].first;
		cnt[A[i].second]++;
			
		if (A[i].first >= mx1) {
			mx2 = mx1;
			mx1 = A[i].first;
		}
		else if (A[i].first >= mx2) {
			mx2 = A[i].first;
		}
	}
	
	bool found = false;
	for (int i = 0; i < n; i++) {
		if (cnt[i] == 2) {
			found = true;
			break;
		}
	}
	
	if (found) {
		ans = min(ans, sum);
	}
	else {		
		for (int i = 0; i < n; i++) {
			if (A[i].first == mx1) {
				ans = min(ans, sum + s[A[i].second] - A[i].first- mx2);
			}
			else {
				ans = min(ans, sum + s[A[i].second] - A[i].first - mx1);
			}
		}
	}
	
	cout << ans;
}

```