Friends and Subsequences

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121333#problem/H

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 max(al_ar)==min(blbr) 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.

题意：

分别已知a b数组任意区间的最大值、最小值；

求有多少区间[l,r]满足max(al_ar)==min(blbr);

题解：

RMQ：O(nlgn)预处理 O(1)求出任意区间的min/max；

在固定区间左端点情况下：

由于最大值最小值均具有单调性；

用两次二分操作分别求出第一次和最后一次满足min==max的右端点，作差累加即可；

注意：两次二分操作的差别和意义.

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <queue>

#include <map>

#include <set>

#include <vector>

#define LL long long

#define eps 1e-8

#define maxn 201000

#define mod 1000000007

#define inf 0x3f3f3f3f

#define IN freopen("in.txt","r",stdin);

using namespace std;

int n, A[maxn], B[maxn];

int d_min[maxn][30];

int d_max[maxn][30];

void RMQ_init() {

    for(int i=0; i<n; i++) d_max[i][0] = A[i], d_min[i][0] = B[i];

    for(int j=1; (1<<j)<=n; j++)

        for(int i=0; i+(1<<j)-1<n; i++) {

            d_min[i][j] = min(d_min[i][j-1], d_min[i+(1<<(j-1))][j-1]);

            d_max[i][j] = max(d_max[i][j-1], d_max[i+(1<<(j-1))][j-1]);

        }

}

int RMQ_min(int L, int R) {

    int k = 0;

    while((1<<(k+1)) <= R-L+1) k++;

    return min(d_min[L][k], d_min[R-(1<<k)+1][k]);

}

int RMQ_max(int L, int R) {

    int k = 0;

    while((1<<(k+1)) <= R-L+1) k++;

    return max(d_max[L][k], d_max[R-(1<<k)+1][k]);

}

int main(int argc, char const *argv[])

{

    //IN;

    while(scanf("%d",&n) != EOF)

    {

        for(int i=0; i<n; i++) scanf("%d",&A[i]);

        for(int i=0; i<n; i++) scanf("%d",&B[i]);

        RMQ_init();

        LL ans = 0;

        for(int i=0; i<n; i++) {

            if(A[i] > B[i]) continue;

            int first_r=-1, last_r=-1;

            int l=i,r=n-1,mid;

            while(l <= r) {

                mid = (l+r) / 2;

                if(RMQ_max(i,mid) == RMQ_min(i,mid)) first_r = mid;

                if(RMQ_max(i,mid) >= RMQ_min(i,mid)) r = mid-1;

                else l = mid+1;

            }

            if(first_r == -1) continue;

            l=i; r=n-1;

            while(l <= r) {

                mid = (l+r) / 2;

                if(RMQ_max(i,mid) > RMQ_min(i,mid))

                    r = mid-1;

                else l = mid+1, last_r = mid;

            }

            ans += last_r - first_r + 1;

        }

        printf("%I64d\n", ans);

    }

    return 0;

}