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++ (GCC 9.2.1) (judge verdict: Runtime Error)

/// IN THE NAME OF GUITAR
#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long int ll;
typedef pair<ll, ll> pll;

#define sz(x)                       (ll) x.size()
#define all(x)                      (x).begin(),(x).end()
#define F                           first
#define S                           second
#define lc                          id << 1
#define rc                          lc | 1

ll Pow(ll a, ll b, ll md, ll ans = 1) {
    for (; b; b >>= 1, a = a * a % md)
        if (b & 1)
            ans = ans * a % md;
    return ans % md;
}
 
const ll MAXN = 5e3 + 20;
const ll INF = 8e18;
const ll MOD = 1e9 + 7;
ll dp[MAXN], pd[2][MAXN], X[MAXN], n, ridi = 0;
vector<ll> adj[MAXN];

void DFS(ll v) {
	dp[v] = INF;
	for (ll u : adj[v]) {
		DFS(u);
	}
	pd[0][0] = 0;
	for (ll i = 1; i <= sz(adj[v]); i++) {
		ll u = adj[v][i - 1];
		for (ll j = MAXN - X[u]; j >= 0; j--) {
			pd[i & 1][j + X[u]] = min(pd[i & 1][j + X[u]], pd[1 - (i & 1)][j] + dp[u]);
		}
		for (ll j = MAXN - dp[u]; j >= 0; j--) {
			pd[i & 1][j + dp[u]] = min(pd[i & 1][j + dp[u]], pd[1 - (i & 1)][j] + X[u]);
		}

	}
	for (ll i = 0; i <= X[v]; i++) {
		dp[v] = min(dp[v], pd[sz(adj[v]) & 1][i]);
	}
	for (ll i = 0; i < 2; i++) {
		for (ll j = 0; j < MAXN; j++) pd[i][j] = INF;
	}
	if (dp[v] >= INF / 10) ridi = 1;
}

int main() {
	scanf("%lld", &n);
	for (ll i = 2; i <= n; i++) {
		ll p; scanf("%lld", &p);
		adj[p].push_back(i);
	}
	for (ll i = 1; i <= n; i++) scanf("%lld", &X[i]);
	DFS(1);
	if (ridi) printf("IMPOSSIBLE\n");
	else printf("POSSIBLE\n");
    return 0;
}