#include <bits/stdc++.h>
using namespace std;
int n, t[105], f[105];
bool g[105][105];
int main() {
  bool fl = true;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> t[i];
    t[i]--;
  }
  for (int i = 0; i < n; i++) {
    cin >> f[i];
    g[i][i] = true;
    if (i - f[i] >= 0) g[i][i - f[i]] = g[i - f[i]][i] = true;
    if (i + f[i] < n) g[i][i + f[i]] = g[i + f[i]][i] = true;
  }
  for (int k = 0; k < n; k++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (g[i][k] && g[k][j]) g[i][j] = true;
      }
    }
  }
  for (int i = 0; i < n; i++)
    if (!g[i][t[i]]) fl = false;
  if (fl)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int n, t[105], f[105];
bool g[105][105];
int main() {
  bool fl = true;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> t[i];
    t[i]--;
  }
  for (int i = 0; i < n; i++) {
    cin >> f[i];
    g[i][i] = true;
    if (i - f[i] >= 0) g[i][i - f[i]] = g[i - f[i]][i] = true;
    if (i + f[i] < n) g[i][i + f[i]] = g[i + f[i]][i] = true;
  }
  for (int k = 0; k < n; k++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (g[i][k] && g[k][j]) g[i][j] = true;
      }
    }
  }
  for (int i = 0; i < n; i++)
    if (!g[i][t[i]]) fl = false;
  if (fl)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

```