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;
void dout() { cerr << endl; }
template <typename Head, typename... Tail>
void dout(Head H, Tail... T) {
  cerr << H << ' ';
  dout(T...);
}
long long MOD = 1000000000 + 7;
int color[100000];
vector<int> g[100000];
int n, p;
long long dp[100000][2];
int was[100000][2];
long long pw(long long a, long long p) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    long long ans = pw(a, p / 2);
    return ans * ans % MOD;
  }
  return a * pw(a, p - 1) % MOD;
}
long long inv(long long a) { return pw(a, MOD - 2); }
long long get(int v, int was_b, int p) {
  if (was[v][was_b]) return dp[v][was_b];
  was[v][was_b] = 1;
  if (color[v] == 1 && !was_b) return 0;
  if (color[v] == 1 || was_b == 0) {
    long long ans = 0;
    long long product = 1;
    for (int i = 0; i < g[v].size(); ++i) {
      if (g[v][i] == p) continue;
      product *= get(g[v][i], 0, v) + get(g[v][i], 1, v);
      product %= MOD;
    }
    ans = product;
    return dp[v][was_b] = ans;
  }
  if (was_b) {
    long long ans = 0;
    long long all = 1;
    for (int i = 0; i < g[v].size(); ++i) {
      if (g[v][i] == p) continue;
      all *= get(g[v][i], 0, v) + get(g[v][i], 1, v);
      all %= MOD;
    }
    for (int i = 0; i < g[v].size(); ++i) {
      if (g[v][i] == p) continue;
      long long prod =
          (all * inv(get(g[v][i], 1, v) + get(g[v][i], 0, v)) % MOD) *
          get(g[v][i], 1, v) % MOD;
      ans += prod;
      ans %= MOD;
    }
    return dp[v][was_b] = ans;
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d", &p);
    g[i + 1].push_back(p), g[p].push_back(i + 1);
  }
  for (int i = 0; i < n; ++i) scanf("%d", &color[i]);
  printf("%I64d", get(0, 1, -1));
  return 0;
}
