#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10,maxn=5050;
vector<int> G[N];
int w[N],f[N],c[N]; 	//dp[i]指i点另外一种颜色的最小值 
const int INF=0x3f3f3f3f;
int dp[2][maxn];
#define y x^1
void dfs(int k){
	for(int i=G[k].size()-1;i>=0;i--)
		dfs( G[k][i] );
	memset(dp,0x3f,sizeof(dp));
	int x=1;dp[x][0]=0;
	for(int i=G[k].size()-1;i>=0;i--){
		int u=G[k][i];
		x^=1;
		memset(dp[x],0x3f,sizeof(dp[x]));
		for(int j=w[k];j>=0;j--){
			if(j>=w[u]) dp[x][j]=min(dp[x][j],dp[y][j-w[u]]+f[u]);
			if(j>=f[u]) dp[x][j]=min(dp[x][j],dp[y][j-f[u]]+w[u]);
		}
	}
	f[k]=INF;
	for(int i=0;i<=w[k];i++)
		f[k]=min(f[k],dp[x][i]);
}
int main()
{
	int n;cin>>n;
	for(int a,i=2;i<=n;i++){
		scanf("%d",&a);
		G[a].push_back(i);
	}
	for(int i=1;i<=n;i++)	scanf("%d",&w[i]);
	dfs(1); 
	if(f[1]<INF)	cout<<"POSSIBLE";
	else cout<<"IMPOSSIBLE";
	return 0;
}
### Prompt

Your task is to create a Cpp solution to 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=1e3+10,maxn=5050;
vector<int> G[N];
int w[N],f[N],c[N]; 	//dp[i]指i点另外一种颜色的最小值 
const int INF=0x3f3f3f3f;
int dp[2][maxn];
#define y x^1
void dfs(int k){
	for(int i=G[k].size()-1;i>=0;i--)
		dfs( G[k][i] );
	memset(dp,0x3f,sizeof(dp));
	int x=1;dp[x][0]=0;
	for(int i=G[k].size()-1;i>=0;i--){
		int u=G[k][i];
		x^=1;
		memset(dp[x],0x3f,sizeof(dp[x]));
		for(int j=w[k];j>=0;j--){
			if(j>=w[u]) dp[x][j]=min(dp[x][j],dp[y][j-w[u]]+f[u]);
			if(j>=f[u]) dp[x][j]=min(dp[x][j],dp[y][j-f[u]]+w[u]);
		}
	}
	f[k]=INF;
	for(int i=0;i<=w[k];i++)
		f[k]=min(f[k],dp[x][i]);
}
int main()
{
	int n;cin>>n;
	for(int a,i=2;i<=n;i++){
		scanf("%d",&a);
		G[a].push_back(i);
	}
	for(int i=1;i<=n;i++)	scanf("%d",&w[i]);
	dfs(1); 
	if(f[1]<INF)	cout<<"POSSIBLE";
	else cout<<"IMPOSSIBLE";
	return 0;
}
```