#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a[200009];
ll b[200009];
ll su[200009];
int main(){
    ll n, s1 = 0, s2 = 0;
    cin >> n;
    vector<ll> S;
    for(ll i = 1; i <= n; i++){
        cin >> a[i] >> b[i];
        S.push_back(a[i]);
        S.push_back(b[i]);
        s1 += a[i];
        s2 += b[i];
    }
    sort(S.begin(), S.end());
    for(ll i = 1; i <= 2*n; i++)
        su[i] = su[i - 1] + S[i-1];
    ll ans =min(s1, s2);
    for(ll i = 1; i <= n; i++){
        ll k = 0, t = 0;
        if(a[i] <= S[n-1])
            {k ++, t += a[i];
            if(b[i] <= S[n])
                k++, t += b[i];
            }
            else
         if(b[i] <= S[n-1])
            {k ++, t += b[i];
            if(a[i] <= S[n])
                k++, t += a[i];
            }
        ans = min(ans, su[n + k] - t);

    }
    cout << ans << endl;
}

### Prompt

Please formulate 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>
#define ll long long
using namespace std;
ll a[200009];
ll b[200009];
ll su[200009];
int main(){
    ll n, s1 = 0, s2 = 0;
    cin >> n;
    vector<ll> S;
    for(ll i = 1; i <= n; i++){
        cin >> a[i] >> b[i];
        S.push_back(a[i]);
        S.push_back(b[i]);
        s1 += a[i];
        s2 += b[i];
    }
    sort(S.begin(), S.end());
    for(ll i = 1; i <= 2*n; i++)
        su[i] = su[i - 1] + S[i-1];
    ll ans =min(s1, s2);
    for(ll i = 1; i <= n; i++){
        ll k = 0, t = 0;
        if(a[i] <= S[n-1])
            {k ++, t += a[i];
            if(b[i] <= S[n])
                k++, t += b[i];
            }
            else
         if(b[i] <= S[n-1])
            {k ++, t += b[i];
            if(a[i] <= S[n])
                k++, t += a[i];
            }
        ans = min(ans, su[n + k] - t);

    }
    cout << ans << endl;
}

```