#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 100005;
struct ii {
	ll a, b;
	bool operator < (ii o) const {
		return tie(a, b) < tie(o.a, o.b);
	}
};

ll N;
ii arr[2*MAXN], pos[MAXN];

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	cin >> N;
	ll a = 0, b = 0;
	for (int i = 0; i < N; i++) {
		cin >> arr[2*i].a >> arr[2*i|1].a;
		pos[i] = {arr[2*i].a, arr[2*i|1].a};
		a += arr[2*i].a;
		b += arr[2*i|1].a;
		arr[2*i].b = arr[2*i|1].b = i;
	}
	sort(arr, arr+2*N);
	
	ll pasti = 0;
	for (int i = 0; i < N-2; i++) pasti += arr[i].a;
	
	ll ans = min(a, b);
	for (int i = 0;i < N; i++) {
		ll cur = 0, idx = N-2;
		if (arr[idx].a >= pos[i].a) {
			cur += arr[idx++].a;
		} else {
			cur += pos[i].a;
		}
		if (arr[idx].a >= pos[i].b) {
			cur += arr[idx++].a;
		} else {
			cur += pos[i].b;
		}
		ans = min(ans, pasti+cur);
	}
	cout << ans << "\n";
	
	return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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 MAXN = 100005;
struct ii {
	ll a, b;
	bool operator < (ii o) const {
		return tie(a, b) < tie(o.a, o.b);
	}
};

ll N;
ii arr[2*MAXN], pos[MAXN];

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	cin >> N;
	ll a = 0, b = 0;
	for (int i = 0; i < N; i++) {
		cin >> arr[2*i].a >> arr[2*i|1].a;
		pos[i] = {arr[2*i].a, arr[2*i|1].a};
		a += arr[2*i].a;
		b += arr[2*i|1].a;
		arr[2*i].b = arr[2*i|1].b = i;
	}
	sort(arr, arr+2*N);
	
	ll pasti = 0;
	for (int i = 0; i < N-2; i++) pasti += arr[i].a;
	
	ll ans = min(a, b);
	for (int i = 0;i < N; i++) {
		ll cur = 0, idx = N-2;
		if (arr[idx].a >= pos[i].a) {
			cur += arr[idx++].a;
		} else {
			cur += pos[i].a;
		}
		if (arr[idx].a >= pos[i].b) {
			cur += arr[idx++].a;
		} else {
			cur += pos[i].b;
		}
		ans = min(ans, pasti+cur);
	}
	cout << ans << "\n";
	
	return 0;
}

```