#include <bits/stdc++.h>
using namespace std;
long double eps = 1e-9;
const long double INF = 1e18;
vector<vector<int> > g;
void add(int a, int b) {
  g[a].push_back(b);
  g[b].push_back(a);
}
vector<bool> used;
bool dfs(int v, int t) {
  used[v] = true;
  if (v == t) return true;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (!used[to]) {
      if (dfs(to, t)) {
        return true;
      }
    }
  }
  return used[t];
}
int main() {
  int n;
  cin >> n;
  g.resize(n);
  vector<int> a;
  vector<int> num;
  int elem;
  for (int i = 0; i < n; i++) {
    cin >> elem;
    a.push_back(elem - 1);
  }
  for (int i = 0; i < n; i++) {
    cin >> elem;
    num.push_back(elem);
  }
  for (int i = 0; i < n; i++) {
    if (i + num[i] < n) {
      add(i, i + num[i]);
    }
    if (i - num[i] >= 0) {
      add(i, i - num[i]);
    }
  }
  for (int i = 0; i < n; i++) {
    used.assign(n, false);
    if (!dfs(i, a[i])) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
long double eps = 1e-9;
const long double INF = 1e18;
vector<vector<int> > g;
void add(int a, int b) {
  g[a].push_back(b);
  g[b].push_back(a);
}
vector<bool> used;
bool dfs(int v, int t) {
  used[v] = true;
  if (v == t) return true;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (!used[to]) {
      if (dfs(to, t)) {
        return true;
      }
    }
  }
  return used[t];
}
int main() {
  int n;
  cin >> n;
  g.resize(n);
  vector<int> a;
  vector<int> num;
  int elem;
  for (int i = 0; i < n; i++) {
    cin >> elem;
    a.push_back(elem - 1);
  }
  for (int i = 0; i < n; i++) {
    cin >> elem;
    num.push_back(elem);
  }
  for (int i = 0; i < n; i++) {
    if (i + num[i] < n) {
      add(i, i + num[i]);
    }
    if (i - num[i] >= 0) {
      add(i, i - num[i]);
    }
  }
  for (int i = 0; i < n; i++) {
    used.assign(n, false);
    if (!dfs(i, a[i])) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  return 0;
}

```