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

typedef long long ll;
const int N = 1e5 + 10;

int n, a[N], b[N];

int main() {
    ll sumA = 0, sumB = 0;
    cin >> n;

    int best = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i] >> b[i];
        sumA += a[i];
        sumB += b[i];
        if (min(a[i], b[i]) > min(a[best], b[best]))
            best = i;
    }
    auto ans = min(sumA, sumB);

    vector<int> pick;
    for (int i = 0; i < n; i++) if (i != best) {
        pick.emplace_back(a[i]); 
        pick.emplace_back(b[i]);
    }
    sort(pick.begin(), pick.end());

    ll sum = 0;
    for (int i = 0; i < int(pick.size()); i++) {
        sum += pick[i];
        if (i+1 == n-2) ans = min(ans, sum + a[best] + b[best]); 
        if (i+1 == n) ans = min(ans, sum);
    }

    cout << ans << endl;
    return 0;
}

### Prompt

Please create a solution in cpp 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 N = 1e5 + 10;

int n, a[N], b[N];

int main() {
    ll sumA = 0, sumB = 0;
    cin >> n;

    int best = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i] >> b[i];
        sumA += a[i];
        sumB += b[i];
        if (min(a[i], b[i]) > min(a[best], b[best]))
            best = i;
    }
    auto ans = min(sumA, sumB);

    vector<int> pick;
    for (int i = 0; i < n; i++) if (i != best) {
        pick.emplace_back(a[i]); 
        pick.emplace_back(b[i]);
    }
    sort(pick.begin(), pick.end());

    ll sum = 0;
    for (int i = 0; i < int(pick.size()); i++) {
        sum += pick[i];
        if (i+1 == n-2) ans = min(ans, sum + a[best] + b[best]); 
        if (i+1 == n) ans = min(ans, sum);
    }

    cout << ans << endl;
    return 0;
}

```