#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <math.h>
using namespace std;

typedef long long ll;
const int MAXN = 2e5 + 10;
int N;
int A[MAXN], B[MAXN];
ll D[MAXN], ta, tb, ans;

int main() {
	register int i, t = 0;
	scanf("%d", &N);
	for(i = 1; i <= N; ++i)
		scanf("%d%d", A + i, B + i),
		ta += A[i], tb += B[i],
		D[++t] = A[i], D[++t] = B[i];
	sort(D + 1, D + (N << 1) + 1);
	ans = min(ta, tb);
	for(i = 1; i <= N << 1; ++i)
		D[i] += D[i - 1];
	for(i = 1; i <= N; ++i) {
		int pos = N; ll res = 0;
		if(A[i] <= D[pos - 1] - D[pos - 2]) ++pos; else res += A[i];
		if(B[i] <= D[pos - 1] - D[pos - 2]) ++pos; else res += B[i];
		ans = min(ans, res + D[pos - 2]);
	}
	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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <math.h>
using namespace std;

typedef long long ll;
const int MAXN = 2e5 + 10;
int N;
int A[MAXN], B[MAXN];
ll D[MAXN], ta, tb, ans;

int main() {
	register int i, t = 0;
	scanf("%d", &N);
	for(i = 1; i <= N; ++i)
		scanf("%d%d", A + i, B + i),
		ta += A[i], tb += B[i],
		D[++t] = A[i], D[++t] = B[i];
	sort(D + 1, D + (N << 1) + 1);
	ans = min(ta, tb);
	for(i = 1; i <= N << 1; ++i)
		D[i] += D[i - 1];
	for(i = 1; i <= N; ++i) {
		int pos = N; ll res = 0;
		if(A[i] <= D[pos - 1] - D[pos - 2]) ++pos; else res += A[i];
		if(B[i] <= D[pos - 1] - D[pos - 2]) ++pos; else res += B[i];
		ans = min(ans, res + D[pos - 2]);
	}
	printf("%lld\n", ans);
	return 0;
}
```