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
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#define N 400400
#define pii pair<int, int>
using namespace std;

int n, m;
bool vis[N];
pii p[N];
long long sum[N];

int main() {
#ifdef isLOCAL
    freopen("work.in", "r", stdin);
    freopen("work.out", "w", stdout);
#endif

    cin >> n;
    long long s1 = 0, s2 = 0;
    for (int i = 1, u, v; i <= n; ++i) {
        scanf("%d%d", &u, &v);
        s1 += u, s2 += v;
        p[++m] = make_pair(u, i);
        p[++m] = make_pair(v, i);
    }
    sort(p + 1, p + m + 1);
    long long ans = min(s1, s2);
    for (int i = 1; i <= n + 2; ++i)
        sum[i] = sum[i - 1] + p[i].first;

    for (int i = 1; i <= n; ++i) {
        if (vis[p[i].second]) return cout << min(ans, sum[n]) << endl, 0;
        vis[p[i].second] = 1;
        if (p[i].second == p[n + 1].second) ans = min(ans, sum[n + 2] - p[n + 1].first - p[i].first);
        else ans = min(ans, sum[n + 1] - p[i].first);
    }

    cout << ans << endl;

    return 0;
}
