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

typedef long long ll;
const int maxn = 2e5;
int n, m, a[maxn + 3], b[maxn + 3], p[maxn + 3][3];
ll sum[maxn + 3];

struct node {
	int id, type, val;
	friend bool operator < (const node &a, const node &b) {
		return a.val < b.val;
	}
} x[maxn + 3];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d %d", &a[i], &b[i]);
		x[++m].id = i, x[m].type = 1, x[m].val = a[i];
		x[++m].id = i, x[m].type = 2, x[m].val = b[i];
	}
	ll A = 0, B = 0;
	for (int i = 1; i <= n; i++) {
		A += a[i], B += b[i];
	}
	ll ans = min(A, B);
	sort(x + 1, x + m + 1);
	for (int i = 1; i <= m; i++) {
		p[x[i].id][x[i].type] = i;
		sum[i] = sum[i - 1] + x[i].val;
	}
	for (int i = 1; i <= n; i++) {
		int l = p[i][1], r = p[i][2];
		if (l > r) {
			swap(l, r);
		}
		if (l - 1 >= n - 2) {
			ans = min(ans, sum[n - 2] + x[l].val + x[r].val);
		} else if (r - 2 >= n - 2) {
			ans = min(ans, sum[n - 1] + x[r].val);
		} else {
			ans = min(ans, sum[n]);
		}
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Generate 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 = 2e5;
int n, m, a[maxn + 3], b[maxn + 3], p[maxn + 3][3];
ll sum[maxn + 3];

struct node {
	int id, type, val;
	friend bool operator < (const node &a, const node &b) {
		return a.val < b.val;
	}
} x[maxn + 3];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d %d", &a[i], &b[i]);
		x[++m].id = i, x[m].type = 1, x[m].val = a[i];
		x[++m].id = i, x[m].type = 2, x[m].val = b[i];
	}
	ll A = 0, B = 0;
	for (int i = 1; i <= n; i++) {
		A += a[i], B += b[i];
	}
	ll ans = min(A, B);
	sort(x + 1, x + m + 1);
	for (int i = 1; i <= m; i++) {
		p[x[i].id][x[i].type] = i;
		sum[i] = sum[i - 1] + x[i].val;
	}
	for (int i = 1; i <= n; i++) {
		int l = p[i][1], r = p[i][2];
		if (l > r) {
			swap(l, r);
		}
		if (l - 1 >= n - 2) {
			ans = min(ans, sum[n - 2] + x[l].val + x[r].val);
		} else if (r - 2 >= n - 2) {
			ans = min(ans, sum[n - 1] + x[r].val);
		} else {
			ans = min(ans, sum[n]);
		}
	}
	printf("%lld\n", ans);
	return 0;
}
```