Codeforces Round #485 (Div. 2) Three displays

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 $n$

displays placed along a road, and the $i$-th of them can display a text with font size $s_i$ only. Maria Stepanovna wants to rent such three displays with indices $i<j<k$ that the font size increases if you move along the road in a particular direction. Namely, the condition $s_i<s_j<s_k$

should be held.

The rent cost is for the $i$

-th display is $c_i$

. Please determine the smallest cost Maria Stepanovna should pay.

Input

The first line contains a single integer $n$

($3\le n\le 3000$

) — the number of displays.

The second line contains $n$

integers $s_1,s_2,\dots ,s_n$ ($1\le s_i\le {10}^9$

) — the font sizes on the displays in the order they stand along the road.

The third line contains $n$

integers $c_1,c_2,\dots ,c_n$ ($1\le c_i\le {10}^8$

) — 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 $i<j<k$

such that $s_i<s_j<s_k$

.

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 $1$

, $4$ and $5$, because $s_1<s_4<s_5$ ($2<4<10$), and the rent cost is $40+10+40=90$

.

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

题意

给你带权值的数，选三个递增的数，使权值最小
思路
dp，先找出每个数结尾两个数最小权值，再dp找第三个数

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
ll dp1[3100],dp2[3100];
struct node
{
    ll data,num;
}a[3100];

int main()
{
    ll n ,i ,j;
    while(cin>>n)
    {
        for(i=0;i<n;i++)
            cin>>a[i].data;
        for(i=0;i<n;i++)
            cin>>a[i].num;
        memset(dp1,0,sizeof(dp1));
        memset(dp2,0,sizeof(dp2));
        for(i=0;i<n;i++)
        {
            dp1[i] = inf;
            for(j=0;j<i;j++)
            {
                if(a[j].data<a[i].data) //
                    dp1[i] = min(dp1[i],a[i].num+a[j].num);//两个数权值最小
            }
        }
        for(i=0;i<n;i++)
        {
            dp2[i] = inf;
            for(j=0;j<i;j++)
            {
                if(a[j].data<a[i].data&&dp1[j]<inf) //被选得数必须dp第一次中符合得数
                {
                    dp2[i] = min(dp2[i],dp1[j]+a[i].num);
                }
            }
        }
        ll ans = inf;
        for(i=0;i<n;i++)
        {
            //cout<<dp1[i]<<" "<<dp2[i]<<endl;
            ans = min(ans,dp2[i]);
        }
        if(ans == inf)
            cout<<"-1"<<endl;
        else
            cout<<ans<<endl;
    }
    return 0;
}