U - Three displays

Problem description

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 si 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 si<sj<sk should be held.

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

Input

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

The second line contains n integers s1,s2,…,sn (1≤si≤10^9) — the font sizes on the displays in the order they stand along the road.

The third line contains n integers c1,c2,…,cn (1≤ci≤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 si<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 1, 4 and 5, because s1<s4<s5(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.

解题思路：题目的意思就是找出连续递增的三个数si,sj,sk，使得si<sj<sk，而且使得ci+cj+ck的和最小。作法：从二个数开始循环到倒数第二个数，每循环到当前值sj就向两边遍历查找符合条件的si、sk对应的最小ci,ck值，若是找获得即m一、m2都≠INF，就将三个c值相加再和原来的最小值mincost进行比较替换；最后若是mincost仍是INF，说明找不到这样符合条件的状况，此时输出"-1"。时间复杂度为是O（n²），暴力即过。

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int INF=3e8+5;//这里INF要设置成比3e8大一点，由于三个c值相加有可能恰好为3e8  4 struct NODE{int s,c;}node[3005];
 5 int main(){
 6     int n,m1,m2,mincost=INF;cin>>n;
 7     for(int i=0;i<n;++i)cin>>node[i].s;
 8     for(int i=0;i<n;++i)cin>>node[i].c;
 9     for(int i=1;i<n-1;++i){
10         m1=m2=INF;
11         for(int j=i-1;j>=0;--j)
12             if(node[j].s<node[i].s&&node[j].c<m1)m1=node[j].c;
13         for(int j=i+1;j<n;++j)
14             if(node[j].s>node[i].s&&node[j].c<m2)m2=node[j].c;
15         if(m1!=INF&&m2!=INF)mincost=min(mincost,node[i].c+m1+m2);
16     }
17     if(mincost<INF)cout<<mincost<<endl;
18     else cout<<"-1"<<endl;
19     return 0;
20 }