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>
using namespace std;

#define rep(i,l,r) for(int i=l;i<=r;++i)
#define per(i,r,l) for(int i=r;i>=l;--i)
const int N=1000+5,U=5000+5;
int fa[N],x1[N],x2[N];
vector<int>lk[N];
int dp[U];

int main()
{
	//freopen("1.in","r",stdin);
	int n;
	cin>>n;
	rep(i,2,n){cin>>fa[i];lk[fa[i]].push_back(i);}
	rep(i,1,n)cin>>x1[i];
	per(x,n,1)
	{
		int u=x1[x];
		rep(i,0,u)dp[i]=0;
		for(auto y:lk[x])
		{
			int a=x1[y],b=x2[y];
			if(a<b)swap(a,b);
			per(i,u,a)dp[i]=min(dp[i-a]+b,dp[i-b]+a);
			per(i,a-1,b)dp[i]=dp[i-b]+a;
			per(i,b-1,0)dp[i]=U;
		}
		if(dp[u]>=U)
		{
			puts("IMPOSSIBLE");
			exit(0);
		}
		x2[x]=dp[u];
	}
	puts("POSSIBLE");
}