#include<bits/stdc++.h>
using namespace std;
const int N=1010,M=5010,INF=10000000;
int f[N][M],tot[N],b[N],ne[N],fi[N],minf[N],k,n,x;
void add(int x,int y){
	b[++k]=y; ne[k]=fi[x]; fi[x]=k;
}
void dp(int x){
	for (int k=1; k<=tot[x]; k++) f[x][k]=INF;
	for (int j=fi[x]; j; j=ne[j]){
		dp(b[j]);
		for (int k=tot[x]; k>=0; k--){
			int t1=INF,t2=INF;
			if (k>=tot[b[j]]) t1=f[x][k-tot[b[j]]]+minf[b[j]];//ÏàÍ¬ 
			if (k>=minf[b[j]]) t2=f[x][k-minf[b[j]]]+tot[b[j]];//²»Í¬
			f[x][k]=min(t1,t2);
//			printf("%d %d %d %d\n",x,b[j],k,f[x][k]);
		}
	}
	minf[x]=f[x][0]; for (int i=1; i<=tot[x]; i++) minf[x]=min(minf[x],f[x][i]);
//	printf("hhd:%d %d %d\n",x,tot[x],minf[x]);
	if (minf[x]==INF){
		puts("IMPOSSIBLE");
		exit(0);
	}
}
int main(){
	scanf("%d",&n);
	for (int i=2; i<=n; i++){
		scanf("%d",&x);
		add(x,i);
	}	
	for (int i=1; i<=n; i++) scanf("%d",&tot[i]);
	dp(1);
	puts("POSSIBLE");
}
### Prompt

Create a solution in Cpp for the following problem:

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`.

Examples

Input

3
1 1
4 3 2


Output

POSSIBLE


Input

3
1 2
1 2 3


Output

IMPOSSIBLE


Input

8
1 1 1 3 4 5 5
4 1 6 2 2 1 3 3


Output

POSSIBLE


Input

1

0


Output

POSSIBLE

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
const int N=1010,M=5010,INF=10000000;
int f[N][M],tot[N],b[N],ne[N],fi[N],minf[N],k,n,x;
void add(int x,int y){
	b[++k]=y; ne[k]=fi[x]; fi[x]=k;
}
void dp(int x){
	for (int k=1; k<=tot[x]; k++) f[x][k]=INF;
	for (int j=fi[x]; j; j=ne[j]){
		dp(b[j]);
		for (int k=tot[x]; k>=0; k--){
			int t1=INF,t2=INF;
			if (k>=tot[b[j]]) t1=f[x][k-tot[b[j]]]+minf[b[j]];//ÏàÍ¬ 
			if (k>=minf[b[j]]) t2=f[x][k-minf[b[j]]]+tot[b[j]];//²»Í¬
			f[x][k]=min(t1,t2);
//			printf("%d %d %d %d\n",x,b[j],k,f[x][k]);
		}
	}
	minf[x]=f[x][0]; for (int i=1; i<=tot[x]; i++) minf[x]=min(minf[x],f[x][i]);
//	printf("hhd:%d %d %d\n",x,tot[x],minf[x]);
	if (minf[x]==INF){
		puts("IMPOSSIBLE");
		exit(0);
	}
}
int main(){
	scanf("%d",&n);
	for (int i=2; i<=n; i++){
		scanf("%d",&x);
		add(x,i);
	}	
	for (int i=1; i<=n; i++) scanf("%d",&tot[i]);
	dp(1);
	puts("POSSIBLE");
}
```