#include <bits/stdc++.h>
using namespace std;

using i64 = int64_t;

const i64 MOD = 1e9+7;

const i64 INF = 1e18+7;

signed main(){
	int n;
	cin >> n;
	vector<int> p(n, -1);
	vector<vector<int>> c(n);
	for(int i = 1; i < n; ++i){
		cin >> p[i];
		c[--p[i]].emplace_back(i);
	}
	vector<int> a(n);
	for(auto& x : a)
		cin >> x;


	vector<int> dp(n, -1);

	for(int i = 0; i < n; ++i)
		if(c[i].empty())
			dp[i] = 0;

	function<bool(int)> f = [&](int x){
		bitset<5007> bs;
		bs.set(0);
		int sum = 0;
		for(auto& y : c[x]){
			if(!f(y))
				return false;
			sum += a[y];
			sum += dp[y];
			bitset<5007> nex;
			nex |= bs << a[y];
			nex |= bs << dp[y];
			bs = move(nex);
		}
		int ret = MOD;
		// 制約を満たす範囲で片側を調べる
		for(int i = 0; i <= a[x]; ++i)
			if(bs[i]){
				// もう片側の値
				int j = sum - i;
				if(j < 0)
					continue;
				if(i <= a[x])
					ret = min(ret, j);
				if(j <= a[x])
					ret = min(ret, i);
			}
		dp[x] = ret;
		return ret != MOD;
	};

	cout << (f(0) ? "" : "IM") << "POSSIBLE" << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;

using i64 = int64_t;

const i64 MOD = 1e9+7;

const i64 INF = 1e18+7;

signed main(){
	int n;
	cin >> n;
	vector<int> p(n, -1);
	vector<vector<int>> c(n);
	for(int i = 1; i < n; ++i){
		cin >> p[i];
		c[--p[i]].emplace_back(i);
	}
	vector<int> a(n);
	for(auto& x : a)
		cin >> x;


	vector<int> dp(n, -1);

	for(int i = 0; i < n; ++i)
		if(c[i].empty())
			dp[i] = 0;

	function<bool(int)> f = [&](int x){
		bitset<5007> bs;
		bs.set(0);
		int sum = 0;
		for(auto& y : c[x]){
			if(!f(y))
				return false;
			sum += a[y];
			sum += dp[y];
			bitset<5007> nex;
			nex |= bs << a[y];
			nex |= bs << dp[y];
			bs = move(nex);
		}
		int ret = MOD;
		// 制約を満たす範囲で片側を調べる
		for(int i = 0; i <= a[x]; ++i)
			if(bs[i]){
				// もう片側の値
				int j = sum - i;
				if(j < 0)
					continue;
				if(i <= a[x])
					ret = min(ret, j);
				if(j <= a[x])
					ret = min(ret, i);
			}
		dp[x] = ret;
		return ret != MOD;
	};

	cout << (f(0) ? "" : "IM") << "POSSIBLE" << endl;
}

```