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 M = 1000000007;
class IO {
  static const int BUF_MB = 1;
  char buf[BUF_MB * 1024 * 1024];
  char *END;
  char *pos;
  int readfile() {
    pos = buf;
    END = fread(buf, 1, sizeof(buf), stdin) + buf;
    return END - buf;
  }

 public:
  IO() { END = pos = buf; }
  int nextInt(int &ret) {
    ret = 0;
    while (true) {
      while (pos < END && !isdigit(*pos)) pos++;
      if (pos == END && readfile() == 0)
        return -1;
      else if (pos != END)
        break;
    }
    while (true) {
      while (pos < END && isdigit(*pos)) {
        ret = ret * 10 + (*pos) - '0';
        pos++;
      }
      if (pos != END || readfile() == 0) break;
    }
    return 1;
  }
} io;
int cl[100005];
int dp[100005][2];
int to[100005];
int main() {
  int n;
  io.nextInt(n);
  for (int i = 0; i < n; i++) {
    dp[i][0] = 1;
    dp[i][1] = 0;
  }
  for (int i = 1; i < n; i++) {
    io.nextInt(to[i]);
  }
  for (int i = 0; i < n; i++) {
    io.nextInt(cl[i]);
  }
  for (int i = n - 1; i >= 0; --i) {
    if (cl[i] == 1) {
      dp[i][1] = dp[i][0];
      dp[i][0] = 0;
    }
    if (i == 0) {
      continue;
    }
    int nxt = to[i];
    dp[nxt][1] = (1ll * dp[nxt][1] * (dp[i][1] + dp[i][0]) +
                  1ll * dp[nxt][0] * dp[i][1]) %
                 M;
    dp[nxt][0] = 1ll * dp[nxt][0] * (dp[i][1] + dp[i][0]) % M;
  }
  cout << dp[0][1] << "\n";
  return 0;
}
