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

uint_fast64_t N, A, B, tap[101010], cnt[4], ans;
vector<tuple<uint_fast64_t, size_t, bool>> p;
bool flg;

int main(){
    scanf("%lld", &N);
    for(size_t i = 0; i < N; ++i){
        scanf("%lld%lld", &A, &B);
        p.emplace_back(A, i, true);
        p.emplace_back(B, i, false);
    }
    sort(p.begin(), p.end());
    for(size_t j = 0; j < N; ++j){
        uint_fast64_t x;
        bool f;
        size_t i;
        tie(x, i, f) = p[j];
        tap[i] += (1 << f);
        ans += x;
    }
    for(size_t i = 0; i < N; ++i)cnt[tap[i]] += 1;
    if(!(cnt[3] || ((!!cnt[1]) ^ (!!cnt[2])))){
        if(get<1>(p[N]) != get<1>(p[N - 1]) || cnt[tap[get<1>(p[N - 1])]] == 1)ans += get<0>(p[N]) - get<0>(p[N - 1]);
        else ans += min(get<0>(p[N]) - get<0>(p[N - 2]), get<0>(p[N + 1]) - get<0>(p[N - 1]));
    }
    printf("%lld\n", ans);
    return 0;
}

### Prompt

Your task is to create 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<bits/stdc++.h>
using namespace std;

uint_fast64_t N, A, B, tap[101010], cnt[4], ans;
vector<tuple<uint_fast64_t, size_t, bool>> p;
bool flg;

int main(){
    scanf("%lld", &N);
    for(size_t i = 0; i < N; ++i){
        scanf("%lld%lld", &A, &B);
        p.emplace_back(A, i, true);
        p.emplace_back(B, i, false);
    }
    sort(p.begin(), p.end());
    for(size_t j = 0; j < N; ++j){
        uint_fast64_t x;
        bool f;
        size_t i;
        tie(x, i, f) = p[j];
        tap[i] += (1 << f);
        ans += x;
    }
    for(size_t i = 0; i < N; ++i)cnt[tap[i]] += 1;
    if(!(cnt[3] || ((!!cnt[1]) ^ (!!cnt[2])))){
        if(get<1>(p[N]) != get<1>(p[N - 1]) || cnt[tap[get<1>(p[N - 1])]] == 1)ans += get<0>(p[N]) - get<0>(p[N - 1]);
        else ans += min(get<0>(p[N]) - get<0>(p[N - 2]), get<0>(p[N + 1]) - get<0>(p[N - 1]));
    }
    printf("%lld\n", ans);
    return 0;
}

```