Three displays

It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.

There are nn displays placed along a road, and the ii-th of them can display a text with font size sisi only. Maria Stepanovna wants to rent such three displays with indices i<j<ki<j<k that the font size increases if you move along the road in a particular direction. Namely, the condition si<sj<sksi<sj<sk should be held.

The rent cost is for the ii-th display is cici. Please determine the smallest cost Maria Stepanovna should pay.

Input

The first line contains a single integer nn (3≤n≤30003≤n≤3000) — the number of displays.

The second line contains nn integers s1,s2,…,sns1,s2,…,sn (1≤si≤1091≤si≤109) — the font sizes on the displays in the order they stand along the road.

The third line contains nn integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤1081≤ci≤108) — the rent costs for each display.

Output

If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices i<j<ki<j<k such that si<sj<sksi<sj<sk.

Examples

input

Copy

5
2 4 5 4 10
40 30 20 10 40

output

Copy

90

input

Copy

3
100 101 100
2 4 5

output

Copy

-1

input

Copy

10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13

output

Copy

33

Note

In the first example you can, for example, choose displays 11, 44 and 55, because s1<s4<s5s1<s4<s5 (2<4<102<4<10), and the rent cost is 40+10+40=9040+10+40=90.

In the second example you can't select a valid triple of indices, so the answer is -1.

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
//#include <xfunctional>
#define ll long long
#define mod 1000000007
using namespace std;
int dir[4][2] = { {0,1},{0,-1},{-1,0},{1,0} };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;

int main()
{
    int n;
    cin >> n;
    vector<int> s(n + 1), c(n + 1);
    vector<vector<ll>> dp(n + 1, vector<ll>(3, INF));
    for (int i = 1; i <= n; i++)
    {
        cin >> s[i];
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> c[i];
    }
    for (int i = 1; i <= n; i++)
    {
        dp[i][0] = c[i];//第i个作为第k个选时的最大值
        for (int k = 1; k < 3; k++)
        {
            for (int j =1; j < i; j++)
            {
                if(s[j]<s[i])
                    dp[i][k] = min(dp[i][k], dp[j][k - 1] + c[i]);
            }
        }
    }
    ll ans = INF;
    for (int i = 1; i <= n; i++)
        ans = min(ans, dp[i][2]);
    printf("%d
", ans == INF ? -1 : ans);
    return 0;
}