Problem Description

Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes up with a problem and ask him about it.

Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}: an+1…a2n. Just like always, there are some restrictions on an+1…a2n: for each number ai, you must choose a number bk from {bi}, and it must satisfy ai≤max{aj-j│bk≤j< i}, and any bk can’t be chosen more than once. Apparently, there are a great many possibilities, so you are required to find max{∑2nn+1ai} modulo 109+7 .

Now Steph finds it too hard to solve the problem, please help him.

Input

The input contains no more than 20 test cases. 
For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}. 
1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n.

Output

For each test case, print the answer on one line: max{∑2nn+1ai} modulo 109+7。

Sample Input

4 
8 11 8 5 
3 1 4 2

Sample Output

27

Hint

For the first sample: 
1. Choose 2 from {bi}, then a_2…a_4 are available for a_5, and you can let a_5=a_2-2=9; 
2. Choose 1 from {bi}, then a_1…a_5 are available for a_6, and you can let a_6=a_2-2=9;

题目大意：

输入一个数s，接下来有两个数组m，n，求m数组从s+1到2s和的最大值。数组m从s+1到2s的计算规则m[i]≤max{m[j]-j│n[k]≤j< i}；并且n数组中的数字每个只能取一次。

#include<iostream> #include<cstdio> #include<cmath> #include<algorithm> const int ll = 1e9 + 7; using namespace std; int m[5000010],n[250010],s[5000010]; int main() { long long int a,b,c,d,e,f,h; while(~scanf("%lld",&a)) { f=0;h=0; for(b=0;b<a;b++) { scanf("%lld",&m[b]); } for(d=0;d<a;d++) { scanf("%lld",&n[d]); } sort(n,n+a); for(b=a-1;b>=0;b--) { if((m[b]-b-1)>h) h=(m[b]-b-1); s[b]=h; } e=-1; for(b=0;b<a;b++) { if(s[n[b]-1]>e) s[a+b]=s[n[b]-1]; else s[a+b]=e; if(s[a+b]-a-b-1>e) e=s[a+b]-a-b-1; f=f+s[a+b]; f=f%ll; } cout<<f<<endl; } return 0; }