#include <bits/stdc++.h>

const int MAXN = 1010;
int fa[MAXN], A[MAXN], f[MAXN], g[5010], h[5010], n;
void gmin(int & x, int y) { x > y ? x = y : 0; }
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 2; i <= n; ++i) std::cin >> fa[i];
	for (int i = 1; i <= n; ++i) std::cin >> A[i];
	for (int i = n; i; --i) {
		memset(g, 0x3f, A[i] + 1 << 2); const int I = *g; *g = 0;
		for (int j = i + 1; j <= n; ++j) if (fa[j] == i) {
			int n1 = I, n2 = I;
			for (int k = 0; k <= A[i]; ++k) {
				if (k >= f[j]) gmin(n1, g[k - f[j]]);
				if (k >= A[j]) gmin(n2, g[k - A[j]]);
				h[k] = std::min(std::min(n1 + A[j], n2 + f[j]), I);
			}
			memcpy(g, h, A[i] + 1 << 2);
		}
		f[i] = *std::min_element(g, g + A[i] + 1);
		if (f[i] == I) return std::cout << "IMPOSSIBLE" << std::endl, 0;
	}
	std::cout << "POSSIBLE" << std::endl;
	return 0;
}

### Prompt

Generate 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>

const int MAXN = 1010;
int fa[MAXN], A[MAXN], f[MAXN], g[5010], h[5010], n;
void gmin(int & x, int y) { x > y ? x = y : 0; }
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 2; i <= n; ++i) std::cin >> fa[i];
	for (int i = 1; i <= n; ++i) std::cin >> A[i];
	for (int i = n; i; --i) {
		memset(g, 0x3f, A[i] + 1 << 2); const int I = *g; *g = 0;
		for (int j = i + 1; j <= n; ++j) if (fa[j] == i) {
			int n1 = I, n2 = I;
			for (int k = 0; k <= A[i]; ++k) {
				if (k >= f[j]) gmin(n1, g[k - f[j]]);
				if (k >= A[j]) gmin(n2, g[k - A[j]]);
				h[k] = std::min(std::min(n1 + A[j], n2 + f[j]), I);
			}
			memcpy(g, h, A[i] + 1 << 2);
		}
		f[i] = *std::min_element(g, g + A[i] + 1);
		if (f[i] == I) return std::cout << "IMPOSSIBLE" << std::endl, 0;
	}
	std::cout << "POSSIBLE" << std::endl;
	return 0;
}

```