Appleman has a tree with n vertices. Some of the vertices (at least one) are colored black and other vertices are colored white.

Consider a set consisting of k (0 ≤ k < n) edges of Appleman's tree. If Appleman deletes these edges from the tree, then it will split into (k + 1) parts. Note, that each part will be a tree with colored vertices.

Now Appleman wonders, what is the number of sets splitting the tree in such a way that each resulting part will have exactly one black vertex? Find this number modulo 1000000007 (109 + 7).

Input

The first line contains an integer n (2 ≤ n ≤ 105) — the number of tree vertices. 

The second line contains the description of the tree: n - 1 integers p0, p1, ..., pn - 2 (0 ≤ pi ≤ i). Where pi means that there is an edge connecting vertex (i + 1) of the tree and vertex pi. Consider tree vertices are numbered from 0 to n - 1.

The third line contains the description of the colors of the vertices: n integers x0, x1, ..., xn - 1 (xi is either 0 or 1). If xi is equal to 1, vertex i is colored black. Otherwise, vertex i is colored white.

Output

Output a single integer — the number of ways to split the tree modulo 1000000007 (109 + 7).

Examples

Input

3
0 0
0 1 1


Output

2


Input

6
0 1 1 0 4
1 1 0 0 1 0


Output

1


Input

10
0 1 2 1 4 4 4 0 8
0 0 0 1 0 1 1 0 0 1


Output

27
#include <bits/stdc++.h>
using namespace std;
struct Edge {
  int np;
  Edge *next;
  pair<long long, long long> val;
} E[101000 * 2], *V[101000];
int tope = -1;
int root;
int c[101000];
void addedge(int x, int y) {
  E[++tope].np = y;
  E[tope].next = V[x];
  V[x] = &E[tope];
}
int fa[101000];
bool nr[101000];
pair<long long, long long> dfs(int now) {
  Edge *ne;
  pair<long long, long long> ret = make_pair(1, 1), pr;
  for (ne = V[now]; ne; ne = ne->next) {
    if (ne->np == fa[now]) continue;
    pr = dfs(ne->np);
    ne->val = pr;
  }
  if (c[now] == 1) {
    ret = make_pair(1, 0);
    for (ne = V[now]; ne; ne = ne->next) {
      if (ne->np == fa[now]) continue;
      ret.first = ret.first * (ne->val.first + ne->val.second) % 1000000007;
    }
    ret.second = 0;
    return ret;
  } else {
    ret = make_pair(0, 1);
    for (ne = V[now]; ne; ne = ne->next) {
      if (ne->np == fa[now]) continue;
      ret.first = (ret.first * (ne->val.second + ne->val.first) % 1000000007 +
                   ret.second * ne->val.first % 1000000007) %
                  1000000007;
      ret.second = ret.second * (ne->val.second + ne->val.first) % 1000000007;
    }
  }
  return ret;
}
int main() {
  int i, j, k, x, y, z;
  int n, m;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%d", &x);
    addedge(x, i + 1);
    addedge(i + 1, x);
    fa[i + 1] = x;
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &c[i]);
  }
  root = 0;
  fa[0] = 0;
  pair<long long, long long> pr;
  long long ans = dfs(root).first % 1000000007;
  printf(
      "%I64d"
      "\n",
      ans);
}
