#include<cstdio>
#include<algorithm>
#include<vector>
#define ll long long
using namespace std;
vector<pair<ll, ll>> vt;
ll check[100010];
int main()
{
	ll n;
	scanf("%lld", &n);
	ll cnt = 0, cnt1 = 0;
	for (int i = 1; i <= n; i++) {
		ll q, w;
		scanf("%lld %lld", &q, &w);
		vt.push_back({ q,i });
		vt.push_back({ w,i });
		cnt += q;
		cnt1 += w;
	}
	sort(vt.begin(), vt.end());
	ll dap = min(cnt, cnt1);
	ll ch = 0;
	int update = 0;
	for (int i = 0; i < n; i++) {
		ch += vt[i].first;
		if (check[vt[i].second])update = 1;
		else check[vt[i].second] = 1;
	}
	if (!update) {
		if (vt[n - 1].second == vt[n].second) {
			ll xx = -vt[n - 1].first + vt[n + 1].first;
			ll yy = vt[n].first - vt[n - 2].first;
			ch += min(xx, yy);
		}
		else {
			ch += (vt[n].first - vt[n - 1].first);
		}
	}
	dap = min(dap, ch);
	printf("%lld\n", dap);
}
### Prompt

Your challenge is to write 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<cstdio>
#include<algorithm>
#include<vector>
#define ll long long
using namespace std;
vector<pair<ll, ll>> vt;
ll check[100010];
int main()
{
	ll n;
	scanf("%lld", &n);
	ll cnt = 0, cnt1 = 0;
	for (int i = 1; i <= n; i++) {
		ll q, w;
		scanf("%lld %lld", &q, &w);
		vt.push_back({ q,i });
		vt.push_back({ w,i });
		cnt += q;
		cnt1 += w;
	}
	sort(vt.begin(), vt.end());
	ll dap = min(cnt, cnt1);
	ll ch = 0;
	int update = 0;
	for (int i = 0; i < n; i++) {
		ch += vt[i].first;
		if (check[vt[i].second])update = 1;
		else check[vt[i].second] = 1;
	}
	if (!update) {
		if (vt[n - 1].second == vt[n].second) {
			ll xx = -vt[n - 1].first + vt[n + 1].first;
			ll yy = vt[n].first - vt[n - 2].first;
			ch += min(xx, yy);
		}
		else {
			ch += (vt[n].first - vt[n - 1].first);
		}
	}
	dap = min(dap, ch);
	printf("%lld\n", dap);
}
```