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.

Example
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.

題意：
給定兩個等長區間，求相同l,r情況下，有多少個區間，第一個序列的最大值和第二個序列的最小值相同。

因為最大值是一個單調遞增的函式，最小值是一個單調遞減的函式，所以 i區間到mid區間的 （最大值-最小值） 是一個單調遞增的函式。所以列舉左端點，不斷的進行二分右端點，分別進行兩個二分可以得到兩個右端點，第一個右端點，代表的是滿足，max(i,mid)==min(i,mid) 的右邊界最大值，然後第二個右端點表示的是max(i,mid)==min(i,mid) 的右邊界最小值。 兩者一減，那麼就是左邊界為i，右邊界在（t1~t2)滿足，這些是都滿足的區間數。
獻上程式碼很好理解。。

無敵快的程式碼

#include <bits/stdc++.h>
using namespace std;
int n;
int a[201000],b[201000];
int mn[201000][30];
int mx[201000][30];
int num;
void init()
{
    for(int i=1;i<=n;i++)
    {
        mx[i][0]=a[i],mn[i][0]=b[i];
    }
    for(int j = 1; (1<<j) <= n; ++j)  
        for(int i = 1; i + (1<<j) - 1 <= n; ++i)  
        {  
            mx[i][j] = max(mx[i][j-1],mx[i+(1<<(j-1))][j-1]);  
            mn[i][j] = min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]);  
        }  
}



int rmq(int i,int j,int f)
{
    int k=0;
      while((1 << (k + 1)) <= j - i + 1) k++;  
    int res=0;
    if(f==0)
        res=min(mn[i][k],mn[j-(1<<k)+1][k]);
    if(f==1)
        res=max(mx[i][k],mx[j-(1<<k)+1][k]);
    return res;
}



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]);
    init();
    long long ans=0;
    for(int i=1;i<=n;i++)
    {
        int l=i,r=n;
        int t1=-1,t2=-1;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            int re1=rmq(i,mid,1),re2=rmq(i,mid,0);
            if(re1>re2)
            {
                r=mid-1;
            }
            else if(re1<re2)
            {
                l=mid+1;
            }
            else if(re1==re2){
                t1=mid;
                r=mid-1;
            }
        }
        if(t1==-1) continue;
        l=i,r=n;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            int re1=rmq(i,mid,1),re2=rmq(i,mid,0);
            if(re1>re2)
                r=mid-1;
            else if(re1<re2)
                l=mid+1;
            else 
            {
                t2=mid;
                l=mid+1;
            }
        }
        ans+=t2-t1+1;
    }
    printf("%lld\n",ans );
}