#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 100010
#define ll long long

template<typename tp> inline void read(tp &x){
    x=0;register char c=getchar();register bool f=0;
    for(;c<'0'||c>'9';f|=(c=='-'),c=getchar());
    for(;c>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0',c=getchar());
    x=f?-x:x;
}
int d[N << 1];
int n;
ll ans = 0;
int main() {
    read(n);
    int a, b;
    read(a), read(b);
    ll ans2 = a, ans3 = b;
    for (int i = 1; i < n; ++i) {
        read(d[i << 1]);
        ans2 += d[i << 1];
        read(d[(i << 1) - 1]);
        ans3 += d[(i << 1) - 1];
        if (max(d[i << 1], d[(i << 1) - 1]) < max(a, b)) {
            swap(a, d[i << 1]);
            swap(b, d[(i << 1) - 1]);
        }
    }
    sort(d + 1, d + n * 2 - 1);
    for (int i = 1; i < n - 1; ++i) ans += d[i];
    ans += min(a + b, d[n] + d[n-1]);
    printf("%lld\n", min(ans, min(ans2, ans3)));
    return 0;
}
### Prompt

Create a solution in Cpp for 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 <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 100010
#define ll long long

template<typename tp> inline void read(tp &x){
    x=0;register char c=getchar();register bool f=0;
    for(;c<'0'||c>'9';f|=(c=='-'),c=getchar());
    for(;c>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0',c=getchar());
    x=f?-x:x;
}
int d[N << 1];
int n;
ll ans = 0;
int main() {
    read(n);
    int a, b;
    read(a), read(b);
    ll ans2 = a, ans3 = b;
    for (int i = 1; i < n; ++i) {
        read(d[i << 1]);
        ans2 += d[i << 1];
        read(d[(i << 1) - 1]);
        ans3 += d[(i << 1) - 1];
        if (max(d[i << 1], d[(i << 1) - 1]) < max(a, b)) {
            swap(a, d[i << 1]);
            swap(b, d[(i << 1) - 1]);
        }
    }
    sort(d + 1, d + n * 2 - 1);
    for (int i = 1; i < n - 1; ++i) ans += d[i];
    ans += min(a + b, d[n] + d[n-1]);
    printf("%lld\n", min(ans, min(ans2, ans3)));
    return 0;
}
```