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 mod = 1000000007, inf = 1000000000;
const int N = 100010;
inline int read() {
  char ch;
  while (!isdigit(ch = getchar()))
    ;
  int x = ch - '0';
  while (isdigit(ch = getchar())) x = (x << 1) + (x << 3) + ch - '0';
  return x;
}
int n, cnt;
int a[N], last[N];
long long f[N][2];
struct edge {
  int to, next;
} e[N << 1];
void insert(int u, int v) {
  e[++cnt].to = v;
  e[cnt].next = last[u];
  last[u] = cnt;
  e[++cnt].to = u;
  e[cnt].next = last[v];
  last[v] = cnt;
}
void dfs(int x, int fa) {
  long long t1, t2;
  if (a[x] == 1) f[x][1] = 1;
  f[x][0] = f[x][1] ^ 1;
  for (int i = last[x]; i; i = e[i].next)
    if (e[i].to != fa) {
      int y = e[i].to;
      dfs(y, x);
      t1 = f[x][0] * (f[y][0] + f[y][1]);
      t2 = (f[x][1] * (f[y][0] + f[y][1]) + f[x][0] * f[y][1]);
      f[x][0] = t1 % mod;
      f[x][1] = t2 % mod;
    }
}
int main() {
  n = read();
  for (int i = 1; i < n; i++) insert(read(), i);
  for (int i = 0; i < n; i++) a[i] = read();
  dfs(0, 0);
  cout << f[0][1] << endl;
  return 0;
}
