#include <bits/stdc++.h>
using namespace std;
const int maxn = 100 + 10;
int par[maxn], d[maxn], a[maxn], b[maxn], n;
map<int, int> m;
int find(int x) {
  int ans = x;
  while (ans != par[ans]) ans = par[ans];
  while (x != par[x]) {
    int y = x;
    x = par[x];
    par[y] = ans;
  }
  return ans;
}
void unn(int x, int y) {
  if (find(x) == find(y)) return;
  if (d[find(x)] > d[find(y)]) swap(x, y);
  int root_x = find(x), root_y = find(y);
  d[root_y] += d[root_x];
  par[root_x] = root_y;
}
void proc() {
  for (int i = 0; i < maxn; i++) {
    par[i] = i;
    d[i] = 1;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> b[i];
  proc();
  for (int i = 1; i <= n; i++) {
    if (i - b[i] > 0) unn(i, i - b[i]);
    if (i + b[i] <= n) unn(i, i + b[i]);
  }
  for (int i = 1; i <= n; i++)
    if (find(i) != find(a[i])) {
      cout << "NO";
      return 0;
    }
  cout << "YES";
}

### Prompt

Please formulate a CPP solution 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;
const int maxn = 100 + 10;
int par[maxn], d[maxn], a[maxn], b[maxn], n;
map<int, int> m;
int find(int x) {
  int ans = x;
  while (ans != par[ans]) ans = par[ans];
  while (x != par[x]) {
    int y = x;
    x = par[x];
    par[y] = ans;
  }
  return ans;
}
void unn(int x, int y) {
  if (find(x) == find(y)) return;
  if (d[find(x)] > d[find(y)]) swap(x, y);
  int root_x = find(x), root_y = find(y);
  d[root_y] += d[root_x];
  par[root_x] = root_y;
}
void proc() {
  for (int i = 0; i < maxn; i++) {
    par[i] = i;
    d[i] = 1;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> b[i];
  proc();
  for (int i = 1; i <= n; i++) {
    if (i - b[i] > 0) unn(i, i - b[i]);
    if (i + b[i] <= n) unn(i, i + b[i]);
  }
  for (int i = 1; i <= n; i++)
    if (find(i) != find(a[i])) {
      cout << "NO";
      return 0;
    }
  cout << "YES";
}

```