#include<bits/stdc++.h>

using ll = long long;

int const N = 200005;

int main() {
	int n;
	std::cin >> n;
	std::vector<std::pair<int, int> > a(n);
	std::vector<int> tot;
	for(int i = 0; i < n; ++i) {
		std::cin >> a[i].first >> a[i].second;
		tot.push_back(a[i].first);
		tot.push_back(a[i].second);
	}
	ll sum1 = 0, sum2 = 0;
	for(auto k : a) {
		sum1 += k.first;
		sum2 += k.second;
	}
	ll ans = std::min(sum1, sum2);
	std::sort(tot.begin(), tot.end());
	ll sum = 0;
	for(int i = 0; i < n; ++i) 
		sum += tot[i];
	for(int i = 0; i < n; ++i) {
		int mx = std::max(a[i].first, a[i].second);
		int mi = std::min(a[i].first, a[i].second);
		if(mi > tot[n - 1] || mx < tot[n - 1]) {
			std::cout << sum << '\n';
			return 0;
		}

		ll tmp = sum;
		tmp += mx - tot[n - 1] + std::max(0, mi - tot[n - 2]);
		ans = std::min(ans, tmp);
	}
	std::cout << ans << '\n';
	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<bits/stdc++.h>

using ll = long long;

int const N = 200005;

int main() {
	int n;
	std::cin >> n;
	std::vector<std::pair<int, int> > a(n);
	std::vector<int> tot;
	for(int i = 0; i < n; ++i) {
		std::cin >> a[i].first >> a[i].second;
		tot.push_back(a[i].first);
		tot.push_back(a[i].second);
	}
	ll sum1 = 0, sum2 = 0;
	for(auto k : a) {
		sum1 += k.first;
		sum2 += k.second;
	}
	ll ans = std::min(sum1, sum2);
	std::sort(tot.begin(), tot.end());
	ll sum = 0;
	for(int i = 0; i < n; ++i) 
		sum += tot[i];
	for(int i = 0; i < n; ++i) {
		int mx = std::max(a[i].first, a[i].second);
		int mi = std::min(a[i].first, a[i].second);
		if(mi > tot[n - 1] || mx < tot[n - 1]) {
			std::cout << sum << '\n';
			return 0;
		}

		ll tmp = sum;
		tmp += mx - tot[n - 1] + std::max(0, mi - tot[n - 2]);
		ans = std::min(ans, tmp);
	}
	std::cout << ans << '\n';
	return 0;
}

```