#include <bits/stdc++.h>
using namespace std;
void initialize(int a[], int n) {
  for (int i = 0; i <= n; i++) a[i] = i;
}
bool find(int a[], int a1, int b1) {
  if (a[a1] == a[b1])
    return true;
  else
    return false;
}
void unionn(int a[], int n, int a1, int b1) {
  int t = a[a1];
  for (int i = 0; i <= n; i++) {
    if (a[i] == t) a[i] = a[b1];
  }
}
int main() {
  int n;
  cin >> n;
  int val[101];
  int fin[101];
  for (int i = 1; i <= n; i++) cin >> fin[i];
  for (int j = 1; j <= n; j++) {
    cin >> val[j];
  }
  val[0] = 0;
  int a[n + 1];
  initialize(a, n);
  for (int i = 1; i <= n; i++) {
    int x = val[i] + i;
    int y = i - val[i];
    if (x <= n) unionn(a, n, a[i], a[x]);
    if (y >= 1) unionn(a, n, a[i], a[y]);
  }
  bool flag = true;
  for (int i = 1; i <= n; i++) {
    if (!find(a, fin[i], i)) flag = false;
  }
  if (flag)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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 - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.

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.

Examples

Input

5
5 4 3 2 1
1 1 1 1 1


Output

YES


Input

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


Output

NO


Input

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


Output

YES

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void initialize(int a[], int n) {
  for (int i = 0; i <= n; i++) a[i] = i;
}
bool find(int a[], int a1, int b1) {
  if (a[a1] == a[b1])
    return true;
  else
    return false;
}
void unionn(int a[], int n, int a1, int b1) {
  int t = a[a1];
  for (int i = 0; i <= n; i++) {
    if (a[i] == t) a[i] = a[b1];
  }
}
int main() {
  int n;
  cin >> n;
  int val[101];
  int fin[101];
  for (int i = 1; i <= n; i++) cin >> fin[i];
  for (int j = 1; j <= n; j++) {
    cin >> val[j];
  }
  val[0] = 0;
  int a[n + 1];
  initialize(a, n);
  for (int i = 1; i <= n; i++) {
    int x = val[i] + i;
    int y = i - val[i];
    if (x <= n) unionn(a, n, a[i], a[x]);
    if (y >= 1) unionn(a, n, a[i], a[y]);
  }
  bool flag = true;
  for (int i = 1; i <= n; i++) {
    if (!find(a, fin[i], i)) flag = false;
  }
  if (flag)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

```