Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3
1 1
4 3 2

Sample Output 1

POSSIBLE

For example, the following allocation satisfies the condition:

Set the color of Vertex 1 to white and its weight to 2.

Set the color of Vertex 2 to black and its weight to 3.

Set the color of Vertex 3 to white and its weight to 2.

There are also other possible allocations.

Sample Input 2

3
1 2
1 2 3

Sample Output 2

IMPOSSIBLE

If the same color is allocated to Vertex 2 and Vertex 3, Vertex 2 cannot be allocated a non-negative weight.

If different colors are allocated to Vertex 2 and 3, no matter which color is allocated to Vertex 1, it cannot be allocated a non-negative weight.

Thus, there exists no allocation of colors and weights that satisfies the condition.

Sample Input 3

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

Sample Output 3

POSSIBLE

Sample Input 4

1

0

Sample Output 4

POSSIBLE

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

//#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define IN(a, b, x) (a<=x&&x<b)
#define MP make_pair
#define PB push_back
const int INF = (1LL<<30);
const ll LLINF = (1LL<<60);
const double PI = 3.14159265359;
const double EPS = 1e-12;
const int MOD = 1000000007;
//#define int ll

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

VI g[5010];

int dp[5010], dp2[5010][5010], x[5010];
void dfs(int num) {
  // 葉なら
  if(g[num].size() == 0) {
    dp[num] = 0;
    return;
  }
  for(int i: g[num]) dfs(i);
  // DPする
  memset(dp2, 0x3f, sizeof(dp2));
  dp2[0][0] = 0;
  REP(i, g[num].size()) REP(j, 5001) {
    if(j>=x[g[num][i]])
      chmin(dp2[i+1][j], dp2[i][j-x[g[num][i]]] + dp[g[num][i]]);
    if(j>=dp[g[num][i]])
      chmin(dp2[i+1][j], dp2[i][j-dp[g[num][i]]] + x[g[num][i]]);
  }
  // cout << "num:" << num << endl;
  // REP(i, g[num].size()+1) {
  //   REP(j, x[num]+1) {
  //     cout << dp2[i][j] << " ";
  //   }
  //   cout << endl;
  // }
  int ret = 0x3f3f3f3f;
  REP(i, x[num]+1) chmin(ret, dp2[g[num].size()][i]);
  if(ret == 0x3f3f3f3f) {
    cout << "IMPOSSIBLE" << endl;
    exit(-1);
  }
  // cout << "num:" << num << " " << ret << endl;
  dp[num] = ret;
}

signed main(void)
{
  int n;
  cin >> n;
  REP(i, n-1) {
    int p;
    cin >> p;
    p--;
    g[p].PB(i+1);
  }
  REP(i, n) cin >> x[i];

  dfs(0);
  // REP(i, n) cout << dp[i] << " "; cout << endl;
  cout << "POSSIBLE" << endl;

  return 0;
}