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

typedef long long ll;

const int N = 100005;
int n, b[2 * N];
bool vis[N];
ll ans, suma, sumb;
pair<int, int> a[N];

int val(int x) {
	return x > 0 ? a[x].first : a[-x].second;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d%d", &a[i].first, &a[i].second);
		suma += a[i].first, sumb += a[i].second;
		b[2 * i - 1] = i, b[2 * i] = -i;
	}
	ans = min(suma, sumb);
	sort(a + 1, a + 1 + n);
	sort(b + 1, b + 1 + 2 * n, [](int x, int y) { return val(x) < val(y); });
	ll sum = 0;
	for (int i = 1; i <= n; ++i)
		sum += val(b[i]);
	for (int i = 1; i <= n; ++i)
		if (vis[abs(b[i])]) {
			printf("%lld\n", min(ans, sum));
			return 0;
		} else
			vis[abs(b[i])] = true;
	for (int i = n + 1; i <= 2 * n; ++i)
		ans = min(ans, sum - (b[i] + b[n] ? val(b[n]) : val(b[n - 1])) + val(b[i]));
	printf("%lld\n", ans);
	return 0;
}

### Prompt

In CPP, your task is to solve 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 N = 100005;
int n, b[2 * N];
bool vis[N];
ll ans, suma, sumb;
pair<int, int> a[N];

int val(int x) {
	return x > 0 ? a[x].first : a[-x].second;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d%d", &a[i].first, &a[i].second);
		suma += a[i].first, sumb += a[i].second;
		b[2 * i - 1] = i, b[2 * i] = -i;
	}
	ans = min(suma, sumb);
	sort(a + 1, a + 1 + n);
	sort(b + 1, b + 1 + 2 * n, [](int x, int y) { return val(x) < val(y); });
	ll sum = 0;
	for (int i = 1; i <= n; ++i)
		sum += val(b[i]);
	for (int i = 1; i <= n; ++i)
		if (vis[abs(b[i])]) {
			printf("%lld\n", min(ans, sum));
			return 0;
		} else
			vis[abs(b[i])] = true;
	for (int i = n + 1; i <= 2 * n; ++i)
		ans = min(ans, sum - (b[i] + b[n] ? val(b[n]) : val(b[n - 1])) + val(b[i]));
	printf("%lld\n", ans);
	return 0;
}

```