YellowStar is versatile. One day he writes a melody A = [A1, ..., AN ], and he has a standard melody B = [B1, ..., BN ]. YellowStar can split melody into several parts, it can be expressed as: K split position S = [S1, ..., SK], satisfy 1 ≤ S1 < S2 < · · · < SK = N. Melody A and B will be split into K parts:

Two parts of melody are equal while the change of tone is consistent. It can be expressed as:
A = [A1, ..., AM ] equal to B = [B1, ..., BM ] need to satisfy:

Now YellowStar wants each part of his melody A equals to each part of standard melody B. In other words, the following conditions need to be met:

YellowStar also wants the number K of split parts as minimum as possible.

Input is given from Standard Input in the following format:
N
A1 A2 . . . AN
B1 B2 . . . BN
Constraints
1 ≤ N ≤ 105
1 ≤ Ai, Bi ≤ 109
All Ai are distinct and all Bi are distinct.
All inputs are integers.

Print one line denotes the minimal integer K .

5

1 3 2 4 5

4 9 10 11 8

3

#include<bits/stdc++.h>

using namespace std;

#define rep(i,a,n) for(int i=a;i<n;i++)

#define scac(x) scanf("%c",&x)

#define sca(x) scanf("%d",&x)

#define sca2(x,y) scanf("%d%d",&x,&y)

#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define scl(x) scanf("%lld",&x)

#define scl2(x,y) scanf("%lld%lld",&x,&y)

#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)

#define pri(x) printf("%d\n",x)

#define pri2(x,y) printf("%d %d\n",x,y)

#define pri3(x,y,z) printf("%d %d %d\n",x,y,z)

#define prl(x) printf("%lld\n",x)

#define prl2(x,y) printf("%lld %lld\n",x,y)

#define prl3(x,y,z) printf("%lld %lld %lld\n",x,y,z)

#define ll long long

#define LL long long

inline ll read(){ll x=,f=;char ch=getchar();

while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

return x*f;}

#define read read()

#define pb push_back

#define mp make_pair

#define P pair<int,int>

#define PLL pair<ll,ll>

#define PI acos(1.0)

#define eps 1e-6

#define inf 1e17

#define INF 0x3f3f3f3f

#define MOD 998244353

#define mod 1e9+7

#define N 1000005

const int maxn=;

ll a[maxn];

ll b[maxn];

int main()

{

  int n;

  sca(n);

  rep(i,,n)

    scl(a[i]);

  rep(i,,n)

    scl(b[i]);

  ll k = ;

  rep(i,,n)

  {

    if((LL)(a[i]-a[i-])*(LL)(b[i]-b[i-]) > )

      continue;

    else

      k++;

  }

  prl(k+);

}