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;
const long long md = 1000000007;
struct node {
  long long white, black;
} tree[100010];
int type[100010];
int n, fa[100010];
void output() {
  printf("\n-----------------------------\n");
  for (int i = 1; i <= n; ++i) {
    printf("tree[%d]: white=%lld, black=%lld\n", i, tree[i].white,
           tree[i].black);
  }
  printf("\n-----------------------------\n");
  return;
}
void run() {
  for (int i = n; i >= 2; --i) {
    int t = fa[i];
    long long bl = tree[i].black * tree[t].white % md;
    bl =
        (bl + ((tree[i].black + tree[i].white) % md) * tree[t].black % md) % md;
    long long wh =
        (((tree[i].black + tree[i].white) % md) * tree[t].white) % md;
    tree[t].black = bl;
    tree[t].white = wh;
  }
  cout << tree[1].black;
  return;
}
void init() {
  cin >> n;
  memset(type, 0, sizeof(type));
  int top = 0, a, b;
  for (int i = 2; i <= n; ++i) {
    scanf("%d", &fa[i]);
    ++fa[i];
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &type[i]);
    if (type[i]) {
      tree[i].white = 0;
      tree[i].black = 1;
    } else {
      tree[i].white = 1;
      tree[i].black = 0;
    }
  }
  run();
  return;
}
int main() {
  init();
  return 0;
}
