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 p(long long x, long long n) {
  x %= 1000000007ll;
  long long ret = 1;
  while (n > 0) {
    if (n & (1ll)) ret = ret * x % 1000000007ll;
    x = x * x % 1000000007ll;
    n >>= (1ll);
  }
  return ret;
}
int n;
vector<int> node[233333];
int x[233333];
long long dp[233333][3];
void dfs(int u, int fa) {
  int cnt = 0;
  for (int i = 0; i < node[u].size(); i++) {
    if (node[u][i] != fa) dfs(node[u][i], u), cnt++;
  }
  if (cnt == 0) {
    if (x[u] == 0) {
      dp[u][0] = 0;
      dp[u][1] = 1;
      dp[u][2] = 0;
    } else {
      dp[u][0] = 1;
      dp[u][1] = 0;
      dp[u][2] = 1;
    }
    return;
  }
  if (x[u] == 1) {
    dp[u][0] = 1;
    dp[u][1] = 0;
    for (int i = 0; i < node[u].size(); i++) {
      int v = node[u][i];
      if (v == fa) continue;
      dp[u][0] =
          (dp[u][0] * ((dp[v][1] + dp[v][2]) % 1000000007ll)) % 1000000007ll;
    }
    dp[u][2] = dp[u][0];
  } else {
    dp[u][1] = 1;
    dp[u][0] = 0;
    for (int i = 0; i < node[u].size(); i++) {
      int v = node[u][i];
      if (v == fa) continue;
      dp[u][0] = dp[u][0] * (dp[v][1] + dp[v][2]) % 1000000007ll +
                 dp[u][1] * dp[v][0] % 1000000007ll;
      dp[u][0] %= 1000000007ll;
      dp[u][1] = dp[u][1] * (dp[v][1] + dp[v][2]) % 1000000007ll;
    }
    dp[u][2] = dp[u][0];
  }
}
int Main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int p;
    cin >> p;
    node[i + 2].push_back(p + 1);
    node[p + 1].push_back(i + 2);
  }
  for (int i = 1; i <= n; i++) {
    cin >> x[i];
  }
  dfs(1, -1);
  cout << dp[1][2] << endl;
  return 0;
}
int main() { return Main(); }
