#include <bits/stdc++.h>

const int N = 1054, W = 5054, INF = 0x3f3f3f3f;

int n;
int p[N], fc[N], nc[N];
int s[N], f[N];
int dp[2][W], *cur = *dp, *nxt = dp[1];

inline void link(int x, int px) {p[x] = px, nc[x] = fc[px], fc[px] = x;}
inline void down(int &x, const int y) {x > y ? x = y : 0;}

void dfs(int x) {
	int i, y;
	for (y = fc[x]; y; y = nc[y]) dfs(y);
	memset(nxt, 63, sizeof *dp), *nxt = 0;
	for (y = fc[x]; y; y = nc[y]) {
		std::swap(cur, nxt), memset(nxt, 63, sizeof *dp);
		for (i = s[y]; i <= s[x]; ++i) down(nxt[i], cur[i - s[y]] + f[y]);
		for (i = f[y]; i <= s[x]; ++i) down(nxt[i], cur[i - f[y]] + s[y]);
	}
	f[x] = *std::min_element(nxt, nxt + (s[x] + 1));
}

int main() {
	int i, x;
	scanf("%d", &n);
	for (i = 2; i <= n; ++i) scanf("%d", &x), link(i, x);
	for (i = 1; i <= n; ++i) scanf("%d", s + i);
	dfs(1), puts(f[1] >= INF ? "IMPOSSIBLE" : "POSSIBLE");
	return 0;
}
### Prompt

Please create a solution in cpp 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 N = 1054, W = 5054, INF = 0x3f3f3f3f;

int n;
int p[N], fc[N], nc[N];
int s[N], f[N];
int dp[2][W], *cur = *dp, *nxt = dp[1];

inline void link(int x, int px) {p[x] = px, nc[x] = fc[px], fc[px] = x;}
inline void down(int &x, const int y) {x > y ? x = y : 0;}

void dfs(int x) {
	int i, y;
	for (y = fc[x]; y; y = nc[y]) dfs(y);
	memset(nxt, 63, sizeof *dp), *nxt = 0;
	for (y = fc[x]; y; y = nc[y]) {
		std::swap(cur, nxt), memset(nxt, 63, sizeof *dp);
		for (i = s[y]; i <= s[x]; ++i) down(nxt[i], cur[i - s[y]] + f[y]);
		for (i = f[y]; i <= s[x]; ++i) down(nxt[i], cur[i - f[y]] + s[y]);
	}
	f[x] = *std::min_element(nxt, nxt + (s[x] + 1));
}

int main() {
	int i, x;
	scanf("%d", &n);
	for (i = 2; i <= n; ++i) scanf("%d", &x), link(i, x);
	for (i = 1; i <= n; ++i) scanf("%d", s + i);
	dfs(1), puts(f[1] >= INF ? "IMPOSSIBLE" : "POSSIBLE");
	return 0;
}
```