类似最大递增子序列CF９８７Ｃ

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

The rent cost is for the $\mathrm{ii-th\; display\; is cici.\; Please\; determine\; the\; smallest\; cost\; Maria\; Stepanovna\; should\; pay.}$

Input

The first line contains a single integer $\mathrm{nn (3\le n\le 30003\le n\le 3000) —\; the\; number\; of\; displays.}$

The second line contains $\mathrm{nn integers s1,s2,\dots ,sns1,s2,\dots ,sn (1\le si\le 1091\le si\le 109) —\; the\; font\; sizes\; on\; the\; displays\; in\; the\; order\; they\; stand\; along\; the\; road.}$

The third line contains $\mathrm{nn integers c1,c2,\dots ,cnc1,c2,\dots ,cn (1\le ci\le 1081\le ci\le 108) —\; the\; rent\; costs\; for\; each\; display.}$

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 $\mathrm{i<j<ki<j<k such\; that si<sj<sksi<sj<sk.}$

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 $\mathrm{11, 44 and 55,\; because s1<s4<s5s1<s4<s5 (2<4<102<4<10),\; and\; the\; rent\; cost\; is 40+10+40=9040+10+40=90.}$

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

#include <bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
const int maxn = 3e3+5;
int a[maxn],b[maxn];
long long dp[maxn][10]; // dp[i][j]表示的是你现在再第i个之后你挑了第j个的时候的最小 值 
//dp[i][j] = dp[k][j-1] + a[j]; 
int main()
{
	int n;
	scanf("%d",&n);
	for(int i = 0 ; i < n ; i++) scanf("%d",&a[i]);
	for(int i = 0 ; i < n ; i++) scanf("%d",&b[i]);
	memset(dp,inf,sizeof(dp));
	for(int i = 0 ; i < n ; i++) dp[i][0] = b[i]; // 初始话一下， 那么第i个数选0本书的时候的最小值为b[i]  0表示的是我已经选了一本书了 
	for(int k = 0 ; k < 3; k ++)
	{
		for(int i = 1 ; i < n ; i++)
		{
			for(int j = 0 ; j < i ; j++)
			{
				if(a[i] > a[j])
				{
					dp[i][k] = min(dp[j][k-1]+b[i],dp[i][k]);
				}
			}
		}
	}
	long long ans = LLONG_MAX;
	for(int i = 0 ; i < n ; i++)
	{
		ans = min(ans,dp[i][2]);
	}
	if(ans > 3e8) ans = -1;
	printf("%d
",ans);
}