codeforces 361 D - Friends and Subsequences

Time Limit:2000MS     Memory Limit:524288KB     64bit IO Format:%I64d & %I64u

 

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, ..., a_n ( - 10⁹ ≤ a_i ≤ 10⁹) — the sequence a.

The third line contains n integer numbers b₁, b₂, ..., b_n ( - 10⁹ ≤ b_i ≤ 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

Input

6 1 2 3 2 1 4 6 7 1 2 3 2

Output

2

Input

3 3 3 3 1 1 1

Output

0

Hint

The occasions in the first sample case are:

1.l = 4,r = 4 since max{2} = min{2}.

2.l = 4,r = 5 since max{2, 1} = min{2, 3}.

There are no occasions in the second sample case since Mike will answer 3 to any query pair, but !Mike will always answer 1.

 

如Hint所示，a为max[]，b为min[]，查找最大与最小区间的相同区间个数

#include<bits/stdc++.h>
using namespace std;
int n,a[200005],b[200005];
long long ans;
deque<int>x,y;
int main()
{
    scanf("%d",&n);
    for(int i=1; i<=n; i++) scanf("%d",&a[i]);
    for(int i=1; i<=n; i++) scanf("%d",&b[i]);
    for(int i=1,j=1; i<=n; i++)
    {
        while(!x.empty()&&a[x.back()]<=a[i])  x.pop_back();
        while(!y.empty()&&b[y.back()]>=b[i])  y.pop_back();
        x.push_back(i);
        y.push_back(i);
        while(j<=i&&a[x.front()]-b[y.front()]>0)
        {
            j++;
            while(!x.empty()&&x.front()<j) x.pop_front();
            while(!y.empty()&&y.front()<j) y.pop_front();
        }
        if(!x.empty()&&!y.empty()&&a[x.front()]==b[y.front()])
        {
            ans+=min(x.front(),y.front())-j+1;
        }
    }
    printf("%lld",ans);
}