#include <bits/stdc++.h>

using namespace std;

const int INF = 1 << 29;

int N, X[1001];
vector< int > g[1000];

pair< int, int > rec(int idx)
{
  vector< int > dp1(5001, 0);

  for(auto &to : g[idx]) {
    auto get = rec(to);

    vector< int > dp2(5001, INF);
    for(int j = 0; j <= 5000 - get.second; j++) {
      dp2[j + get.second] = min(dp2[j + get.second], dp1[j] + get.first);
    }
    for(int j = 0; j <= 5000 - get.first; j++) {
      dp2[j + get.first] = min(dp2[j + get.first], dp1[j] + get.second);
    }
    dp1.swap(dp2);
  }
  return {X[idx], dp1[X[idx]]};
}

int main()
{

  cin >> N;
  for(int i = 1; i < N; i++) {
    int x;
    cin >> x;
    g[--x].emplace_back(i);
  }
  for(int i = 0; i < N; i++) {
    cin >> X[i];
  }
  if(rec(0).second == INF) {
    cout << "IMPOSSIBLE" << endl;
  } else {
    cout << "POSSIBLE" << endl;
  }
}
### Prompt

Your challenge is to write a cpp solution to the following problem:

We have a tree with N vertices. Vertex 1 is the root of the tree, and the parent of Vertex i (2 \leq i \leq N) is Vertex P_i.

To each vertex in the tree, Snuke will allocate a color, either black or white, and a non-negative integer weight.

Snuke has a favorite integer sequence, X_1, X_2, ..., X_N, so he wants to allocate colors and weights so that the following condition is satisfied for all v.

* The total weight of the vertices with the same color as v among the vertices contained in the subtree whose root is v, is X_v.



Here, the subtree whose root is v is the tree consisting of Vertex v and all of its descendants.

Determine whether it is possible to allocate colors and weights in this way.

Constraints

* 1 \leq N \leq 1 000
* 1 \leq P_i \leq i - 1
* 0 \leq X_i \leq 5 000

Inputs

Input is given from Standard Input in the following format:


N
P_2 P_3 ... P_N
X_1 X_2 ... X_N


Outputs

If it is possible to allocate colors and weights to the vertices so that the condition is satisfied, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.

Examples

Input

3
1 1
4 3 2


Output

POSSIBLE


Input

3
1 2
1 2 3


Output

IMPOSSIBLE


Input

8
1 1 1 3 4 5 5
4 1 6 2 2 1 3 3


Output

POSSIBLE


Input

1

0


Output

POSSIBLE

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

const int INF = 1 << 29;

int N, X[1001];
vector< int > g[1000];

pair< int, int > rec(int idx)
{
  vector< int > dp1(5001, 0);

  for(auto &to : g[idx]) {
    auto get = rec(to);

    vector< int > dp2(5001, INF);
    for(int j = 0; j <= 5000 - get.second; j++) {
      dp2[j + get.second] = min(dp2[j + get.second], dp1[j] + get.first);
    }
    for(int j = 0; j <= 5000 - get.first; j++) {
      dp2[j + get.first] = min(dp2[j + get.first], dp1[j] + get.second);
    }
    dp1.swap(dp2);
  }
  return {X[idx], dp1[X[idx]]};
}

int main()
{

  cin >> N;
  for(int i = 1; i < N; i++) {
    int x;
    cin >> x;
    g[--x].emplace_back(i);
  }
  for(int i = 0; i < N; i++) {
    cin >> X[i];
  }
  if(rec(0).second == INF) {
    cout << "IMPOSSIBLE" << endl;
  } else {
    cout << "POSSIBLE" << endl;
  }
}
```