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

int main(){
    int N;
    vector<vector<int>> V;
    cin >> N;
    for(int i=0; i<N; i++){
        int a, b;
        cin >> a >> b;
        V.push_back({a, i, 0});
        V.push_back({b, i, 1});
    }
    sort(V.begin(), V.end());
    int64_t ans = 0;
    int num[100000] = {0}, side[2] = {0};
    for(int i=0; i<N; i++){
        ans += V[i][0];
        num[V[i][1]]++;
        side[V[i][2]]++;
    }

    auto ok = [&]{
        return side[0] == 0 || side[1] == 0 || *min_element(num, num+N) == 0;
    };

    if(ok()){
        // nop
    }else{
        int m2 = V[N-2][0], m1 = V[N-1][0], M1 = V[N][0], M2 = V[N+1][0];
        num[V[N-1][1]]--;
        side[V[N-1][2]]--;
        num[V[N][1]]++;
        side[V[N][2]]++;
        if(ok()){
            ans += M1-m1;
        }else{
            ans += min(M2-m1, M1-m2);
        }
    }
    cout << ans << endl;
    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;

int main(){
    int N;
    vector<vector<int>> V;
    cin >> N;
    for(int i=0; i<N; i++){
        int a, b;
        cin >> a >> b;
        V.push_back({a, i, 0});
        V.push_back({b, i, 1});
    }
    sort(V.begin(), V.end());
    int64_t ans = 0;
    int num[100000] = {0}, side[2] = {0};
    for(int i=0; i<N; i++){
        ans += V[i][0];
        num[V[i][1]]++;
        side[V[i][2]]++;
    }

    auto ok = [&]{
        return side[0] == 0 || side[1] == 0 || *min_element(num, num+N) == 0;
    };

    if(ok()){
        // nop
    }else{
        int m2 = V[N-2][0], m1 = V[N-1][0], M1 = V[N][0], M2 = V[N+1][0];
        num[V[N-1][1]]--;
        side[V[N-1][2]]--;
        num[V[N][1]]++;
        side[V[N][2]]++;
        if(ok()){
            ans += M1-m1;
        }else{
            ans += min(M2-m1, M1-m2);
        }
    }
    cout << ans << endl;
    return 0;
}

```