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<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#define rep(index,num) for(int index=0;index<num;index++)
#define rep1(index,num) for(int index=1;index<=num;index++)
#define scan(argument) cin>>argument
#define prin(argument) cout<<argument<<endl
#define kaigyo cout<<endl
#define eps 1e-7
#define mp(a1,a2) make_pair(a1,a2)
typedef long long ll;
typedef long double ld;
using namespace std;
typedef pair<ll,ll> pll;
typedef pair<int,int> pint;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<pint> vpint;
typedef vector<pll> vpll;
ll INFl=1e+18+1;
int INF=1e+9+1;
int N;
int P[1001],X[1001];
vint ko[1001];
pint costmin[1001];
bool solve(int from){
	costmin[from]=mp(X[from],0);
	int konum=ko[from].size();
	if(konum==0) return 1;
	//printf("from:%d\n",from+1);
	rep(i,konum){
		int to=ko[from][i];
		if(solve(to)==0){
			//printf("solve(%d)=0\n",to+1);
			return 0;
		}
		else{
			//printf("solve(%d)=1\n",to+1);
		}
	}
	vector<vint > knap(konum+1,vint(X[from]+1,-INF));//knap[i][w]:i番目の子まで読んだ時に、自分と同じ色の子孫の重みの総和がwである時の自分と同じ色で塗った直接の子の重みの総和最大値
	int sum=0;
	knap[0][0]=0;
	rep(i,konum){
		int to=ko[from][i];
		rep(w,X[from]+1){
			if(w>=costmin[to].first&&knap[i][w-costmin[to].first]>=0) knap[i+1][w]=max(knap[i+1][w],knap[i][w-costmin[to].first]+costmin[to].first);
			if(w>=costmin[to].second&&knap[i][w-costmin[to].second]>=0) knap[i+1][w]=max(knap[i+1][w],knap[i][w-costmin[to].second]);
			//printf("from:%d i:%d to:%d costmin:(%d,%d) w:%d knap:%d X[from]:%d\n",from+1,i,to+1,costmin[to].first,costmin[to].second,w,knap[i+1][w],X[from]);
		}
		sum+=costmin[to].first;
	}
	int choosemax=0,canchoosewmax=-1;
	bool canchoose=0;
	rep(w,X[from]+1){
		if(knap[konum][w]>=0){
			canchoose=1;
			canchoosewmax=w;
		}
	}
	if(canchoose) choosemax=knap[konum][canchoosewmax];
	//printf("from:%d choosemax:%d canchoose:%d\n",from+1,choosemax,canchoose);
	if(canchoose==0){
		return 0;
	}
	else{
		costmin[from].second=sum-choosemax;
	}
	//printf("costmin[%d]:(%d,%d)\n",from+1,costmin[from].first,costmin[from].second);
	return 1;
}
int main(){
	scan(N);
	P[0]=-1;
	rep1(i,N-1){
		scan(P[i]);
		P[i]--;
		ko[P[i]].push_back(i);
	}
	rep(i,N){
		scan(X[i]);
	}
	if(solve(0)) prin("POSSIBLE");
	else prin("IMPOSSIBLE");
	return 0;
}