#include <bits/stdc++.h>

const int maxn = 1010, maxm = 5010, oo = 12345678;

int n;
std::vector<int> g[maxn];
int dp[maxm], pd[maxm], x[maxn][2];

void chkmin(int &x, int y) {
	(x > y) && (x = y);
}

void dfs(int u) {
	for (int v: g[u]) {
		dfs(v);
	}
	for (int i = 1; i <= x[u][0]; ++i) {
		dp[i] = oo;
	}
	dp[0] = 0;
	for (int v: g[u]) {
		for (int i = 0; i <= x[u][0]; ++i) {
			pd[i] = oo;
			if(i >= x[v][0]) {
				chkmin(pd[i], dp[i - x[v][0]] + x[v][1]);
			}
			if(i >= x[v][1]) {
				chkmin(pd[i], dp[i - x[v][1]] + x[v][0]);
			}
		}
		memcpy(dp, pd, sizeof(dp[0]) * (x[u][0] + 1));
	}
	x[u][1] = oo;
	for (int i = 0; i <= x[u][0]; ++i) {
		chkmin(x[u][1], dp[i]);
	}
	if(x[u][1] == oo) {
		puts("IMPOSSIBLE");
		exit(0);
	}
//	printf("%d\t%d %d\n", u, x[u][0], x[u][1]);
	return;
}

int main() {
	scanf("%d", &n);
	for (int i = 2; i <= n; ++i) {
		int p;
		scanf("%d", &p);
		g[p].push_back(i);
	}
	for (int i = 1; i <= n; ++i) {
		scanf("%d", x[i]);
	}
	dfs(1);
	puts("POSSIBLE");
	return 0;
}
### Prompt

In cpp, your task is to solve 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, maxm = 5010, oo = 12345678;

int n;
std::vector<int> g[maxn];
int dp[maxm], pd[maxm], x[maxn][2];

void chkmin(int &x, int y) {
	(x > y) && (x = y);
}

void dfs(int u) {
	for (int v: g[u]) {
		dfs(v);
	}
	for (int i = 1; i <= x[u][0]; ++i) {
		dp[i] = oo;
	}
	dp[0] = 0;
	for (int v: g[u]) {
		for (int i = 0; i <= x[u][0]; ++i) {
			pd[i] = oo;
			if(i >= x[v][0]) {
				chkmin(pd[i], dp[i - x[v][0]] + x[v][1]);
			}
			if(i >= x[v][1]) {
				chkmin(pd[i], dp[i - x[v][1]] + x[v][0]);
			}
		}
		memcpy(dp, pd, sizeof(dp[0]) * (x[u][0] + 1));
	}
	x[u][1] = oo;
	for (int i = 0; i <= x[u][0]; ++i) {
		chkmin(x[u][1], dp[i]);
	}
	if(x[u][1] == oo) {
		puts("IMPOSSIBLE");
		exit(0);
	}
//	printf("%d\t%d %d\n", u, x[u][0], x[u][1]);
	return;
}

int main() {
	scanf("%d", &n);
	for (int i = 2; i <= n; ++i) {
		int p;
		scanf("%d", &p);
		g[p].push_back(i);
	}
	for (int i = 1; i <= n; ++i) {
		scanf("%d", x[i]);
	}
	dfs(1);
	puts("POSSIBLE");
	return 0;
}
```