#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<int> P(n), D(n);
  for (int i = 0; i < n; ++i) {
    cin >> P[i];
    P[i]--;
  }
  for (int i = 0; i < n; ++i) cin >> D[i];
  vector<vector<int> > A(n, vector<int>(n, 0));
  for (int i = 0; i < n; ++i) {
    int k1 = i - D[i];
    if (k1 >= 0) A[k1][i] = A[i][k1] = 1;
    int k2 = i + D[i];
    if (k2 < n) A[k2][i] = A[i][k2] = 1;
  }
  for (int k = 0; k < n; ++k)
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        if (A[i][k] == 1 && A[k][j] == 1) A[i][j] = A[j][i] = 1;
  for (int i = 0; i < n; ++i) {
    if (P[i] == i) continue;
    int k = -1;
    for (int j = 0; j < n; ++j)
      if (P[j] == i) {
        k = j;
        break;
      }
    if (A[k][i] == 0) {
      cout << "NO";
      return 0;
    }
    swap(P[k], P[i]);
  }
  cout << "YES";
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 main() {
  int n;
  cin >> n;
  vector<int> P(n), D(n);
  for (int i = 0; i < n; ++i) {
    cin >> P[i];
    P[i]--;
  }
  for (int i = 0; i < n; ++i) cin >> D[i];
  vector<vector<int> > A(n, vector<int>(n, 0));
  for (int i = 0; i < n; ++i) {
    int k1 = i - D[i];
    if (k1 >= 0) A[k1][i] = A[i][k1] = 1;
    int k2 = i + D[i];
    if (k2 < n) A[k2][i] = A[i][k2] = 1;
  }
  for (int k = 0; k < n; ++k)
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        if (A[i][k] == 1 && A[k][j] == 1) A[i][j] = A[j][i] = 1;
  for (int i = 0; i < n; ++i) {
    if (P[i] == i) continue;
    int k = -1;
    for (int j = 0; j < n; ++j)
      if (P[j] == i) {
        k = j;
        break;
      }
    if (A[k][i] == 0) {
      cout << "NO";
      return 0;
    }
    swap(P[k], P[i]);
  }
  cout << "YES";
  return 0;
}

```