#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int Nmax = 1e5 + 5;

int n, i, A, B;
bool ok[Nmax];
vector< pair<int,int> > v;
ll sA, sB, sum, ans;

int main()
{
   /// freopen("input", "r", stdin);
    cin.sync_with_stdio(false);

    cin >> n;
    for(i=0; i<n; ++i)
    {
        cin >> A >> B;
        v.push_back({A, i});
        v.push_back({B, i});
        sA += A; sB += B;
    }

    ans = min(sA, sB);

    sort(v.begin(), v.end());

    for(i=0; i<n; ++i) ok[v[i].second] = 1, sum += v[i].first;

    for(i=0; i<n && ok[i]; ++i);

    if(i != n) ans = min(ans, sum);
        else
        {
            if(v[n-1].second != v[n].second) ans = min(ans, sum + v[n].first - v[n-1].first);
                else ans = min(ans, min(sum + v[n].first - v[n-2].first, sum + v[n+1].first - v[n-1].first));
        }

    cout << ans << '\n';
    return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 Nmax = 1e5 + 5;

int n, i, A, B;
bool ok[Nmax];
vector< pair<int,int> > v;
ll sA, sB, sum, ans;

int main()
{
   /// freopen("input", "r", stdin);
    cin.sync_with_stdio(false);

    cin >> n;
    for(i=0; i<n; ++i)
    {
        cin >> A >> B;
        v.push_back({A, i});
        v.push_back({B, i});
        sA += A; sB += B;
    }

    ans = min(sA, sB);

    sort(v.begin(), v.end());

    for(i=0; i<n; ++i) ok[v[i].second] = 1, sum += v[i].first;

    for(i=0; i<n && ok[i]; ++i);

    if(i != n) ans = min(ans, sum);
        else
        {
            if(v[n-1].second != v[n].second) ans = min(ans, sum + v[n].first - v[n-1].first);
                else ans = min(ans, min(sum + v[n].first - v[n-2].first, sum + v[n+1].first - v[n-1].first));
        }

    cout << ans << '\n';
    return 0;
}

```