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)

//此程序使用了qt专属名称，请勿模仿。侵权必究！！！
//我真是太菜了啊……
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1010;
int n,t=0,b[N<<1],w[N],f[N][2]={},g[N],v[N<<1];
bool flag=0;
struct tq
{
	int y;
	int nxt;
}a[N<<1];
inline void in(int x,int y)
{
	a[++t].nxt=b[x];
	b[x]=t;
	a[t].y=y;
	return;
}
inline int read()
{
	int sum=0,flag=1;
	char c;
	for(;c<'0'||c>'9';c=getchar())
	{
		if(c=='-')
		{
			flag=-1;
		}
	}
	for(;c>='0'&&c<='9';c=getchar())
	{
		sum=(sum<<3)+(sum<<1)+c-'0';
	}
	return sum*flag;
}
inline void dfs(int x,int fa)
{
	int mx=0,mn=0,cnt=0;
	for(int i=b[x];i;i=a[i].nxt)
	{
		int y=a[i].y;
		if(y==fa)
		{
			continue;
		}
		dfs(y,x);
		if(flag)
		{
			return;
		}
		mx+=f[y][0];
		mn+=f[y][1];
		g[++cnt]=f[y][0]-f[y][1];
	}
	if(mx<mn)
	{
		swap(mx,mn);
	}
	if(mn>w[x])
	{
		puts("IMPOSSIBLE");
		flag=1;
		return;
	}
	if(mx<=w[x])
	{
		f[x][0]=w[x];
		f[x][1]=mn;
		return;
	}
	memset(v,0,sizeof(v));
	int m=w[x]-mn;
	v[0]=1;
	for(int i=1;i<=cnt;++i)
	{
		for(int j=m;j>=g[i];--j)
		{
			v[j]+=v[j-g[i]];
		}
	}
	f[x][0]=mx;
	f[x][1]=w[x];
	for(int i=m;i;--i)
	{
		if(v[i])
		{
			f[x][0]=mx-i;
			break;
		}
	}
	if(f[x][0]<f[x][1])
	{
		swap(f[x][0],f[x][1]);
	}
	return;
}
int main()
{
	n=read();
	for(int i=2;i<=n;++i)
	{
		int x=read();
		in(x,i);
		in(i,x);
	}
	for(int i=1;i<=n;++i)
	{
		w[i]=read();
	}
	dfs(1,0);
	if(!flag)
	{
		puts("POSSIBLE");
	}
	return 0;
}