#include <bits/stdc++.h>
using namespace std;
map<pair<int, int>, bool> mem;
bool obl(int a, int b, vector<vector<int> > &dp, vector<bool> &cl) {
  if (a == b) return true;
  for (int i = 0; i < dp.size(); i++)
    if (dp[a][i] == 1 && a != i && !cl[i]) {
      cl[i] = true;
      if (obl(i, b, dp, cl)) return true;
    }
  return false;
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  vector<vector<int> > dp(n, vector<int>(n, 0));
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    if (i - b[i] >= 0) {
      dp[i][i - b[i]] = 1;
      dp[i - b[i]][i] = 1;
    }
    if (i + b[i] < b.size()) {
      dp[i][i + b[i]] = 1;
      dp[i + b[i]][i] = 1;
    }
  }
  bool ok = true;
  vector<bool> cl;
  for (int i = 0; i < n && ok; i++) {
    cl.clear();
    cl.assign(n, 0);
    if (!obl(i, a[i], dp, cl)) ok = false;
  }
  if (ok)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
map<pair<int, int>, bool> mem;
bool obl(int a, int b, vector<vector<int> > &dp, vector<bool> &cl) {
  if (a == b) return true;
  for (int i = 0; i < dp.size(); i++)
    if (dp[a][i] == 1 && a != i && !cl[i]) {
      cl[i] = true;
      if (obl(i, b, dp, cl)) return true;
    }
  return false;
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  vector<vector<int> > dp(n, vector<int>(n, 0));
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    if (i - b[i] >= 0) {
      dp[i][i - b[i]] = 1;
      dp[i - b[i]][i] = 1;
    }
    if (i + b[i] < b.size()) {
      dp[i][i + b[i]] = 1;
      dp[i + b[i]][i] = 1;
    }
  }
  bool ok = true;
  vector<bool> cl;
  for (int i = 0; i < n && ok; i++) {
    cl.clear();
    cl.assign(n, 0);
    if (!obl(i, a[i], dp, cl)) ok = false;
  }
  if (ok)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

```