#include <iostream>
#include <cstdio>
#include <vector>
#include <bitset>
#include <cmath>

using namespace std;

typedef long long ll;

ll n;
ll a[1005], b[1005];
ll d[1005];
vector<ll> ve[1005];
bitset<5005> bs;

int main()
{
    scanf("%lld", &n);
    for(ll i = 2; i <= n; i++)
    {
        scanf("%lld", &a[i]);
        ve[a[i]].push_back(i);
    }
    for(ll i = 1; i <= n; i++)
    {
        scanf("%lld", &b[i]);
    }
    for(ll i = n; i >= 1; i--)
    {
        ll t = b[i];
        bs = 1;
        ll sum = 0;
        ll now = 0;
        for(ll j = 0; j < ve[i].size(); j++)
        {
            t -= min(b[ve[i][j]], d[ve[i][j]]);
            sum += max(b[ve[i][j]], d[ve[i][j]]);
            bs |= (bs << abs(b[ve[i][j]] - d[ve[i][j]]));
        }
        if(t < 0)
        {
            printf("IMPOSSIBLE\n");
            return 0;
        }
        for(ll i = t; i >= 0; i--)
        {
            if(bs[i]) {now = i; break;}
        }
        d[i] = sum - now;
    }
    printf("POSSIBLE\n");
    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 <iostream>
#include <cstdio>
#include <vector>
#include <bitset>
#include <cmath>

using namespace std;

typedef long long ll;

ll n;
ll a[1005], b[1005];
ll d[1005];
vector<ll> ve[1005];
bitset<5005> bs;

int main()
{
    scanf("%lld", &n);
    for(ll i = 2; i <= n; i++)
    {
        scanf("%lld", &a[i]);
        ve[a[i]].push_back(i);
    }
    for(ll i = 1; i <= n; i++)
    {
        scanf("%lld", &b[i]);
    }
    for(ll i = n; i >= 1; i--)
    {
        ll t = b[i];
        bs = 1;
        ll sum = 0;
        ll now = 0;
        for(ll j = 0; j < ve[i].size(); j++)
        {
            t -= min(b[ve[i][j]], d[ve[i][j]]);
            sum += max(b[ve[i][j]], d[ve[i][j]]);
            bs |= (bs << abs(b[ve[i][j]] - d[ve[i][j]]));
        }
        if(t < 0)
        {
            printf("IMPOSSIBLE\n");
            return 0;
        }
        for(ll i = t; i >= 0; i--)
        {
            if(bs[i]) {now = i; break;}
        }
        d[i] = sum - now;
    }
    printf("POSSIBLE\n");
    return 0;
}

```