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 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)

ll n;
ll check[1010];
ll root[1010];
ll x[1010];
ll y[1010];


ll ko[1010];
ll kocnt;
ll dp[5010];
void f(ll k){//頂点kの塗り方を決める

	//子孫の塗り方をチェックする
	check[k]=1;
	rep(i,0,n)if(root[i]==k&&!check[i])f(i);
	
	kocnt=0;
	rep(i,0,n)if(root[i]==k)ko[kocnt++]=i;

	//x[k]以下の最大値を探すdp
	ll upper=x[k];
	rep(i,0,5010)dp[i]=0;
	dp[0]=1;
	rep(i,0,kocnt){
		ll d=x[ko[i]]-y[ko[i]];
		for(ll j=5005;j>=d;j--)if(dp[j-d])dp[j]=1;
		upper-=y[ko[i]];
	}
	if(upper<0){
		puts("IMPOSSIBLE");
		exit(0);
	}
	for(ll j=upper;j>=0;j--)if(dp[j]){
		//j+ΣY=Z
		//求めるのはΣX+ΣY-Z=ΣX-j
		rep(i,0,kocnt)j-=x[ko[i]];
		y[k]=-j;
		break;
	}
}

int main(){
	scanf("%lld",&n);
	root[0]=-1;
	rep(i,1,n){
		ll t;
		scanf("%lld",&t);
		root[i]=t-1;
	}
	rep(i,0,n)scanf("%lld",x+i);
	
	f(0);
	puts("POSSIBLE");
//	rep(i,0,n)printf("%lld ",y[i]);
	return 0;
}