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 int M = 1e5 + 7;
const long long int mod = 1e9 + 7;
const long long int infi = LLONG_MAX;
long long int ans, k, n, x, y, m, mymax = LLONG_MIN, mymin = LLONG_MAX, b, c, z,
                                  sum;
long long int dp[M][2];
long long int color[M], visit[M];
vector<long long int> v[M];
long long int binpow(long long int base, long long int pow,
                     long long int modulus) {
  long long int ans = 1;
  base = base % modulus;
  while (pow > 0) {
    while (pow % 2 == 0) {
      base = (base * base) % modulus;
      pow /= 2;
    }
    ans = (ans * base) % modulus;
    pow--;
  }
  return ans;
}
long long int dfs(long long int node) {
  visit[node] = 1;
  long long int flag = 0;
  long long int val1 = 1, val2 = 0;
  for (auto kk : v[node]) {
    if (visit[kk] != 1) {
      flag = 1;
      dfs(kk);
      val1 = ((val1 * (dp[kk][0] + dp[kk][1])) % mod);
    }
  }
  if (flag == 0) {
    dp[node][color[node]] = 1;
    visit[node] = 0;
    return 0;
  }
  if (color[node] == 1) {
    dp[node][1] = val1;
    visit[node] = 0;
    return 0;
  }
  for (auto kk : v[node]) {
    if (visit[kk] != 1) {
      long long int temp1, temp2;
      temp1 = ((val1 * (dp[kk][1])) % mod);
      temp2 = binpow(dp[kk][1] + dp[kk][0], mod - 2, mod);
      temp1 = ((temp2 * temp1) % mod);
      val2 = (val2 + temp1) % mod;
    }
  }
  dp[node][1] = val2;
  dp[node][0] = val1;
  visit[node] = 0;
}
int main() {
  long long int i, j;
  scanf("%lld", &n);
  for (i = 1; i < n; i++) {
    scanf("%lld", &x);
    v[i].push_back(x);
    v[x].push_back(i);
  }
  for (i = 0; i < n; i++) {
    scanf("%lld", &x);
    if (x == 1) y = i;
    color[i] = x;
  }
  dfs(0);
  printf("%lld\n", dp[0][1]);
  return 0;
}
