D. Friends and Subsequences

题目连接：

Description

Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on their own, but together (with you) — who knows?

Every one of them has an integer sequences a and b of length n. Being given a query of the form of pair of integers (l, r), Mike can instantly tell the value of while !Mike can instantly tell the value of .

Now suppose a robot (you!) asks them all possible different queries of pairs of integers (l, r) (1 ≤ l ≤ r ≤ n) (so he will make exactly n(n + 1) / 2 queries) and counts how many times their answers coincide, thus for how many pairs is satisfied.

How many occasions will the robot count?

Input

The first line contains only integer n (1 ≤ n ≤ 200 000).

The second line contains n integer numbers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the sequence a.

The third line contains n integer numbers b1, b2, ..., bn ( - 109 ≤ bi ≤ 109) — the sequence b.

Output

Print the only integer number — the number of occasions the robot will count, thus for how many pairs is satisfied

Sample Input

6

1 2 3 2 1 4

6 7 1 2 3 2

Sample Output

2

Hint

题意

给你一个a数组和一个b数组

问你有多少对(l,r)满足，a数组中max(L,R)恰好等于b数组中的min(L，R)

题解

暴力枚举L，然后二分相等的那个区间就好了。

因为max肯定是递增的，min是递减的

那个相等的区间可以二分出来。

代码

#include
    
     using namespace std;const int maxn = 2e5+7;int n;int a[maxn],b[maxn];struct RMQ{    const static int RMQ_size = maxn;    int n;    int ArrayMax[RMQ_size][21];    int ArrayMin[RMQ_size][21];    void build_rmq(){        for(int j = 1 ; (1<
     
      <= n ; ++ j)            for(int i = 0 ; i + (1<
      
       b[i])continue;        int l=i,r=n,ansl=i;        while(l<=r){            int mid=(l+r)/2;            if(s1.QueryMax(i,mid)>=s2.QueryMin(i,mid))r=mid-1,ansl=mid;            else l=mid+1;        }        if(s1.QueryMax(i,ansl)>s2.QueryMin(i,ansl))continue;        l=i,r=n;        int ansr=i;        while(l<=r){            int mid=(l+r)/2;            if(s1.QueryMax(i,mid)>s2.QueryMin(i,mid))r=mid-1,ansr=mid;            else l=mid+1;        }        ans+=ansr-ansl;    }    cout<
       
        <