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 <iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<math.h>
using namespace std;
typedef long long ll;
#define int long long
typedef vector<int> VI;
typedef pair<int, int> pii;
typedef priority_queue<int> PQ;
#define fore(i,a) for(auto &i:a)
#define REP(i,n) for(int i=0;i<n;i++)
#define eREP(i,n) for(int i=0;i<=n;i++)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define eFOR(i,a,b) for(int i=(a);i<=(b);++i)
#define SORT(c) sort((c).begin(),(c).end())
#define rSORT(c) sort((c).rbegin(),(c).rend())
#define LB(x,a) lower_bound((x).begin(),(x).end(),(a))
#define UB(x,a) upper_bound((x).begin(),(x).end(),(a))
#define INF 1000000000
#define LLINF 9223372036854775807
#define mod 1000000007
//vector<vector<int> > dp;
//vector<vector<vector<int> > > vvvi;
//dp=vector<vector<int> >(N, vector<int>(M,0));
//vector<pair<int,int> > v;
//v.push_back(make_pair(x,y));
//priority_queue<int,vector<int>, greater<int> > q2;

int N;
vector<VI>tree(1010);
VI X(1010);
string ans = "POSSIBLE";
int dp[1010][2];

void dfs(int cu, int pa = -1) {
	int white = 0, black = 0;
	VI any;
	REP(i, tree[cu].size()) {
		int to = tree[cu][i];
		if (to != pa) {
			dfs(to, cu);

			if (dp[to][0] == INF && dp[to][1] == INF) {
				ans = "IMPOSSIBLE";
				return;
			}
			if (dp[to][0] == INF) {
				white = dp[to][1];
				black = X[to];
			}
			else if (dp[to][1] == INF) {
				white = X[to];
				black = dp[to][1];
			}
			else {
				any.push_back(to);
			}
		}
	}

	VI dp2(5050);
	REP(i, 5050)dp2[i] = INF;
	dp2[white] = black;
	REP(i, any.size()) {
		int ch = any[i];
		VI pd2(5050);
		REP(i, 5050)pd2[i] = INF;
		for (int i = 5000; i >= 0; i--) {
			if (dp2[i] != INF) {
				if (i + X[ch] < 5050)pd2[i + X[ch]] = min(pd2[i + X[ch]], dp2[i] + dp[ch][1]);
				if (i + dp[ch][0] < 5050)pd2[i + dp[ch][0]] = min(pd2[i + dp[ch][0]], dp2[i] + X[ch]);
			}
		}
		swap(dp2, pd2);
	}
	REP(i, X[cu] + 1)if (dp2[i] != INF)dp[cu][0] = min(dp[cu][0], dp2[i]);

	dp2.resize(5050, INF);
	dp2[black] = white;
	REP(i, any.size()) {
		int ch = any[i];
		VI pd2(5050);
		REP(i, 5050)pd2[i] = INF;
		
		for (int i = 5000; i >= 0; i--) {
			if (dp2[i] != INF) {
				if (i + X[ch] < 5050)pd2[i + X[ch]] = min(pd2[i + X[ch]], dp2[i] + dp[ch][0]);
				if (i + dp[ch][1] < 5050)pd2[i + dp[ch][0]] = min(pd2[i + dp[ch][1]], dp2[i] + X[ch]);
			}
		}
		swap(dp2, pd2);
	}
	
	REP(i, X[cu] + 1)if (dp2[i] != INF)dp[cu][1] = min(dp[cu][1], dp2[i]);

}

signed main(){
cin.tie(0);
ios::sync_with_stdio(false);

	cin >> N;
	
	FOR(i, 1, N) {
		int a;
		cin >> a;
		a--;
		tree[a].push_back(i);
		tree[i].push_back(a);
	}
	REP(i, N)cin >> X[i];
	REP(i, N) {
		dp[i][0] = INF;
		dp[i][1] = INF;
	}

	dfs(0);

	if (dp[0][0] == INF && dp[0][1] == INF)ans = "IMPOSSIBLE";

	cout << ans << endl;

	return 0;
}