One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it‘s ordinal number (starting from 1). Also each cell determined it‘s favourite number. On it‘s move i-th cell can exchange it‘s value with the value of some other j-th cell, if |i

The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.

Input

The first line contains positive integer n (1?≤?n?≤?100) — the number of cells in the array. The second line contains n distinct integers from 1 to n — permutation. The last line contains n integers from 1 to n — favourite numbers of the cells.

Output

If the given state is reachable in the described game, output YES, otherwise NO.

Example

5
5 4 3 2 1
1 1 1 1 1

YES

7
4 3 5 1 2 7 6
4 6 6 1 6 6 1

NO

7
4 2 5 1 3 7 6
4 6 6 1 6 6 1

YES

分析：
需要判斷每個數是否能移動到它所指定的位置上，
這裏兩個數經過一些中間量進行位置交換的時候，不影響中間量的位置。
所以可以獨立的判斷每次交換。
這裏我們把能夠互相交換的序號放在同一個集合裏，運用並查集來判斷每個數能否移動到指定位置
代碼如下：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int pre[110];
int n;
int id[110];
int val[110];
int Find(int x)
{
    int h=x,tmp;
    while(pre[x]!=x)
    x=pre[x];
    while(h!=x)
    {
        tmp=pre[h];
        pre[h]=x;
        h=tmp;
    }
    return x;
}
void init()
{
    for(int i=1;i<=n;i++)
    pre[i]=i;
}
void join(int x,int y)
{
    int p=Find(x);
    int q=Find(y);
    if(p!=q)
    pre[p]=q;
}
int main()
{
    ios::sync_with_stdio(false);
    cin>>n;
    init();
    int x,flag;
    flag=1;
    for(int i=1;i<=n;i++)
    {
        cin>>x;
        id[x]=i;
    }
    for(int i=1;i<=n;i++)
    cin>>val[i];
    for(int i=1;i<=n;i++)
    {
      if(i-val[i]>=1)
      join(i,i-val[i]);
      if(i+val[i]<=n)
      join(i,i+val[i]);
    }
    for(int i=1;i<=n;i++)
    {
      if(Find(id[i])!=Find(i))
      {
          flag=0;
          break;
      }
    }
    if(flag==1)puts("YES");
    else puts("NO");
    return 0;
}

Codeforces 28B. pSort （並查集）