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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
typedef pair<int,int> pa;
#define w1 first
#define w2 second
#define ls (x<<1)
#define rs (x<<1|1)
#define pb push_back
#define mp make_pair
#define mid ((l+r)>>1)
#define SZ(x) ((x).size())
#define All(x) (x).begin(),(x).end()
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define rep2(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define per(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)
#define Rep(p,x) for(int (p)=head[(x)];(p);(p)=nxt[(p)])
template<class T>void read(T&num){
	num=0;T f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
	num*=f;
}
int power(int x,int k,int p){int res=1;for(;k;k>>=1,x=1ll*x*x%p)if(k&1)res=1ll*res*x%p;return res;}
int getgcd(int x,int y){return !x?y:getgcd(y%x,x);}
const int maxn=1e3+5,inf=1e9+5;
int n,tot;
int a[maxn];
int head[maxn],des[maxn<<1],nxt[maxn<<1];
int f[maxn][5005],g[maxn][5005];
void adde(int x,int y){
	des[++tot]=y;nxt[tot]=head[x];head[x]=tot;
}
int dfs(int x,int fa){
	int leaf=1;
	rep(i,0,5000)f[x][i]=g[x][i]=inf;
	Rep(p,x)if(des[p]!=fa){
		int num=dfs(des[p],x);
		if(num==inf)return inf;
		if(leaf){
			g[x][num]=a[des[p]];
			g[x][a[des[p]]]=num;
		}else{
			per(i,5000,0){
				if(i>=a[des[p]])g[x][i]=min(g[x][i],f[x][i-a[des[p]]]+num);
				if(i>=num)g[x][i]=min(g[x][i],f[x][i-num]+a[des[p]]);
			}
		}
		rep(i,0,5000)f[x][i]=g[x][i],g[x][i]=inf;
		leaf=0;
	}
	if(leaf)return 0;
	int res=inf;
	rep(i,0,a[x])res=min(res,f[x][i]);
	return res;
}
int main(){
	read(n);
	rep(i,2,n){
		int x;read(x);
		adde(x,i);adde(i,x);
	}
	rep(i,1,n)read(a[i]);
	if(dfs(1,0)==inf)puts("IMPOSSIBLE");
	else puts("POSSIBLE");
	return 0;
}