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 <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAX_N = 1e5 + 10;

int N;
ll A[MAX_N], B[MAX_N], ans = LLONG_MAX;
pair<ll, int> seq[MAX_N * 2];
ll sum[MAX_N];
vector<int> pos[MAX_N];

int main() {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) scanf("%lld%lld", &A[i], &B[i]);
    ans = min(ans, accumulate(A + 1, A + N + 1, 0ll));
    ans = min(ans, accumulate(B + 1, B + N + 1, 0ll));
    for (int i = 1; i <= N; i++) seq[i] = make_pair(A[i], i);
    for (int i = 1; i <= N; i++) seq[N + i] = make_pair(B[i], i);
    sort(seq + 1, seq + 2 * N + 1);
    for (int i = 1; i <= 2 * N; i++) pos[seq[i].second].push_back(i);
    for (int i = 1; i <= 2 * N; i++) sum[i] = sum[i - 1] + seq[i].first;
    for (int i = 1; i <= N; i++) {
        int x = N;
        if (pos[i][0] > x || pos[i][1] > x) x--;
        if (pos[i][0] > x && pos[i][1] > x) x--;
        ll res = sum[x];
        for (int j : pos[i])
            if (j > x) res += seq[j].first;
        ans = min(ans, res);
    }
    printf("%lld\n", ans);
}
