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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> pii;
typedef double db;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod;assert(b>=0);for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head
const int N1=1100;
const int N2=5100;
vi e[N1];
int n,x[N1];
pii best[N1];
vector<vi> dp(N1,vi(N2,0));
bool dfs(int v,int p) {
	int siz=0;
	for(auto a:e[v]) {
		if(a!=p) {
			if(!dfs(a,v)) return 0;
		}
	}
	for(auto a:e[v]) {
		if(a!=p) {
			siz+=best[a].fi+best[a].se;
		}
	}
	dp[v][0]=1;
	for(auto a:e[v]) {
		if(a!=p) {
			per(i,0,x[v]+1) {
				if(dp[v][i]) {
					dp[v][i]=0;
					if(i+best[a].fi<x[v]+1) {
						dp[v][i+best[a].fi]=1;
					}
					if(i+best[a].se<=x[v]) {
						dp[v][i+best[a].se]=1;
					}
				}
			}
		}
	}
	per(i,0,x[v]+1) {
		if(dp[v][i]) {
			best[v].fi=x[v];
			int add=x[v]-i;
			siz+=add;
			best[v].se=siz-x[v];
			//cout<<v<<' '<<best[v].fi<<' '<<best[v].se<<endl;
			return 1;
		}
	}
	return 0;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n;
	rep(i,2,n+1) {
		int a;cin>>a;
		e[a].pb(i);
		e[i].pb(a);
	}
	rep(i,0,n) cin>>x[i+1];
	//cout<<dfs(1,0)<<endl;
	if(dfs(1,0)) puts("POSSIBLE");
	else puts("IMPOSSIBLE");
	return 0;
}