Friends and Subsequences

                               B. Friends and Subsequences                              time limit per test:2 seconds                            memory limit per test:512 megabytes                                    inputstandard: input                                   outputstandard: output

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.

Examples

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

Note

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：给定两个序列A、B，问存在多少个连续子序列，满足max{A}=min{B}？

思路：考虑max{A}<=min{B}的情况，如果满足max{A}<=min{B}，那么这个序列的子序列一定也满足，这样就满足单调性；那么我们就可以枚举区间右端点，去二分左端点，满足max{A}=min{B}的应该是一个区间，所以我们应该分别运用>=和<=去二分出这个区间的左右断点，将差值累加到结果中。这里还需要查询区间最大值，一开始我是用线段树写的，不过好像是被卡常数，超时了，后来改为了可以O(1)查询RMQ，就AC了。

code:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define MAXN 202000#define INF 0x3f3f3f3fint n;int ma[MAXN][20],mi[MAXN][20];void cal(){    for(int j=1;(1<<j)<=n;j++)    {        for(int i=0;i+(1<<j)-1<n;i++)        {            ma[i][j] = max(ma[i][j-1],ma[i+(1<<(j-1))][j-1]);            mi[i][j] = min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);        }    }}int RMQmax(int li,int ri){    int k = 0;    while((1<<(k+1)) <=  ri-li+1) k++;    return max(ma[li][k],ma[ri-(1<<k)+1][k]);}int RMQmin(int li,int ri){    int k = 0;    while((1<<(k+1)) <=  ri-li+1) k++;    return min(mi[li][k],mi[ri-(1<<k)+1][k]);}int main(){    scanf("%d",&n);    for(int i=0;i<n;i++) scanf("%d",&ma[i][0]);    for(int i=0;i<n;i++) scanf("%d",&mi[i][0]);    cal();    long long ans = 0;    int l,r,ma,mb,st,ed;    for(int i=0;i<n;i++)    {        l = i;        r = n-1;        st = -1;        while(l<=r)        {            int mid = (l+r)>>1;            ma = RMQmax(i,mid);            mb = RMQmin(i,mid);            if(ma >= mb)            {                if(ma == mb) st = mid;                r = mid-1;            }            else                l = mid+1;        }        if(st == -1) continue;        l = i;        r = n-1;        ed = -1;        while(l<=r)        {            int mid = (l+r)>>1;            ma = RMQmax(i,mid);            mb = RMQmin(i,mid);            if(ma > mb)                r = mid-1;            else            {                if(ma == mb) ed = mid;                l = mid+1;            }        }        ans += (ed - st + 1);    }    printf("%I64d\n",ans);    return 0;}