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: Wrong Answer)

#include<bits/stdc++.h>
#define FR(i,a,b) for(int i=a;i>=b;i--) 
#define FO(i,a,b) for(int i=a;i<=b;i++)
#define FE(i,a,b) for(int i=a;i<b;i++)
using namespace std;
const int N=1005;
int head[N],to[N],nxt[N],f[N],X[N];
int t,n,possible;
void add(int x,int y) {
	nxt[++t]=head[x],head[x]=t,to[t]=y;
}
int tmp[2][5005],rep;
void dfs(int u) {
	if(head[u]==-1) {
		f[u]=0;
		return ;
	}
	int sum=0;
	for(int i=head[u];i!=-1;i=nxt[i]) 
		dfs(to[i]),sum+=X[to[i]]+f[to[i]];
	memset(tmp,0,sizeof(tmp)),rep=-1;
	for(int i=head[u];i!=-1;i=nxt[i]) {
		rep++;
		if(i==head[u]) {
			if(X[to[i]]<=X[u])
				tmp[rep&1][X[to[i]]]=1;
			if(f[to[i]]<=X[u])
				tmp[rep&1][f[to[i]]]=1;
		}
		else {
			FO(j,f[to[i]],X[u]) if(tmp[(rep-1)&1][j-f[to[i]]]) tmp[rep&1][j]=1;
			FO(j,X[to[i]],X[u])	if(tmp[(rep-1)&1][j-X[to[i]]]) tmp[rep&1][j]=1;
		}
	}
//	if(u==1) 
//		cout<<"FUCK\n";
	FR(i,X[u],0) if(tmp[rep&1][i]) {
		f[u]=sum-i;return ;
	}
	possible=0;
}
int main() {
//	freopen("data.in","r",stdin);
	memset(head,-1,sizeof(head));
	t=-1;
	cin>>n;
	int fa;possible=1;
	FO(i,2,n) cin>>fa,add(fa,i);
	FO(i,1,n) cin>>X[i];
	dfs(1);
	if(possible) cout<<"POSSIBLE";
	else cout<<"IMPOSSIBLE";
	return 0;
}