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;
long long int Pow(long long int a, long long int b, long long int md,
                  long long int ans = 1) {
  for (; b; b >>= 1, a = a * a % md)
    if (b & 1) ans = ans * a % md;
  return ans % md;
}
const long long int MAXN = 1e6 + 10;
const long long int INF = 8e18;
const long long int MOD = 1e9 + 7;
long long int A[MAXN], dp[2][MAXN], n;
vector<long long int> adj[MAXN];
long long int inv(long long int x) { return Pow(x, MOD - 2, MOD); }
void DFS(long long int v) {
  if ((long long int)adj[v].size() == 0) {
    if (A[v])
      dp[0][v] = 1;
    else
      dp[1][v] = 1;
    return;
  }
  long long int sum = 1;
  for (long long int u : adj[v]) {
    DFS(u);
    sum = sum * (dp[0][u] + dp[1][u]) % MOD;
  }
  if (A[v] == 0)
    dp[1][v] = sum;
  else {
    dp[0][v] = sum;
    return;
  }
  for (long long int u : adj[v]) {
    long long int x = sum * inv((dp[0][u] + dp[1][u]) % MOD) % MOD;
    dp[0][v] = (dp[0][v] + x * dp[0][u] % MOD) % MOD;
  }
}
int main() {
  scanf("%lld", &n);
  for (long long int i = 2, p; i <= n; i++) {
    scanf("%lld", &p);
    adj[p + 1].push_back(i);
  }
  for (long long int i = 1; i <= n; i++) scanf("%lld", &A[i]);
  DFS(1);
  printf("%lld\n", dp[0][1]);
  return 0;
}
