Three displays CodeForces - 987C （dp）

C. Three displays

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

5
2 4 5 4 10
40 30 20 10 40

output

90

input

3
100 101 100
2 4 5

output

-1

input

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

output

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.

题意：拿三样物品，价值要和顺序都递增，cost总和最少

题解：dp，改一下板子就可以了

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn=;
 
int main()
{
    int n;
    int ans=INF;
    int val[maxn],cost[maxn],f[maxn][];
    cin>>n;
    for(int i=;i<n;i++)
        cin>>val[i];
    for(int i=;i<n;i++)
        cin>>cost[i];
    memset(f, INF, sizeof(f));
    for (int i = ; i < n; i++)
    {
        f[i][] = cost[i];
        for (int j = ; j < i; j++)
            if (val[j] < val[i])
                f[i][] =min(f[i][], f[j][] + cost[i]);
        for (int j = ; j < i; j++)
            if (val[j] < val[i])
                f[i][] =min(f[i][], f[j][] + cost[i]);
 
    }
    for(int i=;i<n;i++)
        ans=min(ans,f[i][]);
    if(ans!=INF)
        cout<<ans<<endl;
    else
        cout<<-<<endl;
 
}