#include <bits/stdc++.h>
#define rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;

typedef long long ll;
typedef pair<ll, int> pli;
const int N = 2e5 + 10;
int n, a[N], b[N];
ll tot, suma, sumb, sum, res;
pli all[N];
bool vis[N];

int main() {
    cin >> n;
    rep(i, 1, n) {
        scanf("%d%d", &a[i], &b[i]);
        tot += a[i] + b[i];
        suma += a[i], sumb += b[i];
        all[i * 2 - 1] = make_pair(a[i], i);
        all[i * 2] = make_pair(b[i], i);
    }
    sort(all + 1, all + (n << 1) + 1);
    bool ff = 0;
    res = min(suma, sumb);
    rep(i, 1, n) {
        if (vis[all[i].second]) {
            ff = 1;
        }
        vis[all[i].second] = 1;
        sum += all[i].first;
    }
    if (ff) {
        res = min(res, sum);
    } else {
        rep(i, n + 1, n << 1) {
            if (all[n].second == all[i].second)
                res = min(res, sum + all[i].first - all[n - 1].first);
            else
                res = min(res, sum + all[i].first - all[n].first);
        }
    }
    printf("%lld\n", res);
    return 0;
}
### Prompt

In Cpp, your task is to solve 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>
#define rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;

typedef long long ll;
typedef pair<ll, int> pli;
const int N = 2e5 + 10;
int n, a[N], b[N];
ll tot, suma, sumb, sum, res;
pli all[N];
bool vis[N];

int main() {
    cin >> n;
    rep(i, 1, n) {
        scanf("%d%d", &a[i], &b[i]);
        tot += a[i] + b[i];
        suma += a[i], sumb += b[i];
        all[i * 2 - 1] = make_pair(a[i], i);
        all[i * 2] = make_pair(b[i], i);
    }
    sort(all + 1, all + (n << 1) + 1);
    bool ff = 0;
    res = min(suma, sumb);
    rep(i, 1, n) {
        if (vis[all[i].second]) {
            ff = 1;
        }
        vis[all[i].second] = 1;
        sum += all[i].first;
    }
    if (ff) {
        res = min(res, sum);
    } else {
        rep(i, n + 1, n << 1) {
            if (all[n].second == all[i].second)
                res = min(res, sum + all[i].first - all[n - 1].first);
            else
                res = min(res, sum + all[i].first - all[n].first);
        }
    }
    printf("%lld\n", res);
    return 0;
}
```