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;
int pr[1000007];
bool isnotprime(long long x) {
  if (x == 1) return 1;
  if (x == 2) return 0;
  if (x % 2 == 0) return 1;
  return (pr[x / 64] & (1 << ((x >> 1) & 31)));
}
void sieve(long long n) {
  for (long long i = 3; i * i <= n; i += 2) {
    if (!isnotprime(i))
      for (long long j = i * i, k = i << 1; j < n; j += k)
        pr[j / 64] |= (1 << ((j >> 1) & 31));
  }
}
long long fpow(long long x, long long y) {
  x = x % 1000000007;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= 1000000007;
    y = y >> 1;
    x = x * x;
    x %= 1000000007;
  }
  return res;
}
long long inv(long long a, long long m) {
  long long c = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += c;
  return x;
}
bool visited[100005];
vector<long long> adj[100005];
long long col[100005];
long long dp[100005][2];
void dfs(long long v) {
  visited[v] = true;
  dp[v][1] = col[v];
  dp[v][0] = 1 - col[v];
  for (auto u : adj[v]) {
    if (visited[u])
      continue;
    else {
      dfs(u);
      long long zero = dp[v][0];
      long long one = dp[v][1];
      dp[v][0] = 0;
      dp[v][1] = 0;
      dp[v][1] +=
          ((zero * dp[u][1]) % 1000000007 + (one * dp[u][0]) % 1000000007) %
          1000000007;
      dp[v][1] %= 1000000007;
      dp[v][0] += (zero * dp[u][0]) % 1000000007;
      dp[v][0] %= 1000000007;
      dp[v][0] += (zero * dp[u][1]) % 1000000007;
      dp[v][0] %= 1000000007;
      dp[v][1] += (one * dp[u][1]) % 1000000007;
      dp[v][1] %= 1000000007;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    long long x;
    cin >> x;
    adj[i + 1].push_back(x);
    adj[x].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) cin >> col[i];
  dfs(0);
  cout << dp[0][1];
}
