#include <bits/stdc++.h>
using namespace std;
int mat[105][105];
int perm[105];
int fav[105];
int main() {
  int n;
  int i, j, k, tmp;
  for (i = 0; i < 105; i++) {
    for (j = 0; j < 105; j++) {
      mat[i][j] = 200000000;
    }
  }
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &perm[i]);
    perm[i]--;
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &fav[i]);
  }
  for (i = 0; i < n; i++) {
    mat[i][i] = 0;
    if (i + fav[i] < n) {
      mat[i][i + fav[i]] = mat[i + fav[i]][i] = 1;
    }
    if (i - fav[i] > -1) {
      mat[i][i - fav[i]] = mat[i - fav[i]][i] = 1;
    }
  }
  for (k = 0; k < n; k++) {
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if (mat[i][k] + mat[k][j] < mat[i][j])
          mat[i][j] = mat[i][k] + mat[k][j];
      }
    }
  }
  for (i = 0; i < n; i++) {
    if (mat[i][perm[i]] == 200000000 && mat[perm[i]][i] == 200000000) break;
  }
  if (i < n)
    printf("NO\n");
  else
    printf("YES\n");
  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 mat[105][105];
int perm[105];
int fav[105];
int main() {
  int n;
  int i, j, k, tmp;
  for (i = 0; i < 105; i++) {
    for (j = 0; j < 105; j++) {
      mat[i][j] = 200000000;
    }
  }
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &perm[i]);
    perm[i]--;
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &fav[i]);
  }
  for (i = 0; i < n; i++) {
    mat[i][i] = 0;
    if (i + fav[i] < n) {
      mat[i][i + fav[i]] = mat[i + fav[i]][i] = 1;
    }
    if (i - fav[i] > -1) {
      mat[i][i - fav[i]] = mat[i - fav[i]][i] = 1;
    }
  }
  for (k = 0; k < n; k++) {
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if (mat[i][k] + mat[k][j] < mat[i][j])
          mat[i][j] = mat[i][k] + mat[k][j];
      }
    }
  }
  for (i = 0; i < n; i++) {
    if (mat[i][perm[i]] == 200000000 && mat[perm[i]][i] == 200000000) break;
  }
  if (i < n)
    printf("NO\n");
  else
    printf("YES\n");
  return 0;
}

```