#include <bits/stdc++.h>
using namespace std;
long long int GC(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return GC(b, a % b);
}
long long int inv_gc(long long int a, long long int b) {
  return a / GC(a, b) * b;
}
long long int Ceil(long long int a, long long int b) {
  return a / b + (a % b != 0);
}
int par[200];
int rk[200];
void make_set(int n) {
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    rk[i] = 1;
  }
}
int find(int x) {
  if (x == par[x]) return x;
  int p = find(par[x]);
  par[x] = p;
  return p;
}
void uni(int x, int y) {
  int p1 = find(x);
  int p2 = find(y);
  if (p1 == p2) return;
  if (rk[p1] > rk[p2]) {
    par[p2] = p1;
    rk[p1] += rk[p2];
  } else {
    par[p1] = p2;
    rk[p2] += rk[p1];
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int A[200];
  int f[200];
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
  }
  for (int i = 1; i <= n; i++) cin >> f[i];
  make_set(n);
  for (int i = 1; i <= n; i++) {
    if (i + f[i] >= 1 && i + f[i] <= n) uni(i, i + f[i]);
    if (i - f[i] >= 1 && i - f[i] <= n) uni(i, i - f[i]);
  }
  bool k = true;
  for (int i = 1; i <= n; i++) {
    if (find(i) == find(A[i])) {
      continue;
    } else if (find(i) != find(A[i])) {
      k = false;
      break;
    }
  }
  if (k)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  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;
long long int GC(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return GC(b, a % b);
}
long long int inv_gc(long long int a, long long int b) {
  return a / GC(a, b) * b;
}
long long int Ceil(long long int a, long long int b) {
  return a / b + (a % b != 0);
}
int par[200];
int rk[200];
void make_set(int n) {
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    rk[i] = 1;
  }
}
int find(int x) {
  if (x == par[x]) return x;
  int p = find(par[x]);
  par[x] = p;
  return p;
}
void uni(int x, int y) {
  int p1 = find(x);
  int p2 = find(y);
  if (p1 == p2) return;
  if (rk[p1] > rk[p2]) {
    par[p2] = p1;
    rk[p1] += rk[p2];
  } else {
    par[p1] = p2;
    rk[p2] += rk[p1];
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int A[200];
  int f[200];
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
  }
  for (int i = 1; i <= n; i++) cin >> f[i];
  make_set(n);
  for (int i = 1; i <= n; i++) {
    if (i + f[i] >= 1 && i + f[i] <= n) uni(i, i + f[i]);
    if (i - f[i] >= 1 && i - f[i] <= n) uni(i, i - f[i]);
  }
  bool k = true;
  for (int i = 1; i <= n; i++) {
    if (find(i) == find(A[i])) {
      continue;
    } else if (find(i) != find(A[i])) {
      k = false;
      break;
    }
  }
  if (k)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

```