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 int max_n = 111111, inf = 1000000007;
int n, col[max_n];
long long dp[max_n][2];
vector<int> g[max_n];
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  if (b % 2 == 0) {
    return power((a * a) % inf, b / 2);
  }
  return (a * power(a, b - 1)) % inf;
}
void dfs(int v, int p) {
  for (int i = 0; i < g[v].size(); ++i) {
    if (g[v][i] != p) {
      dfs(g[v][i], v);
    }
  }
  if (g[v].size() == 1 && v) {
    dp[v][0] = 1;
    dp[v][1] = 1;
    if (col[v] == 0) {
      dp[v][1] = 0;
    }
    return;
  }
  if (col[v] == 1) {
    dp[v][1] = 1;
    for (int i = 0; i < g[v].size(); ++i) {
      if (g[v][i] != p) {
        dp[v][1] = (dp[v][1] * dp[g[v][i]][0]) % inf;
      }
    }
  } else {
    vector<long long> a, b;
    int f = 0, poz;
    for (int i = 0; i < g[v].size(); ++i) {
      if (g[v][i] != p) {
        a.push_back(dp[g[v][i]][0]);
        b.push_back(dp[g[v][i]][1]);
      }
    }
    for (int i = 0; i < a.size(); ++i) {
      if (a[i] == 0) {
        ++f;
        poz = i;
      }
    }
    if (f >= 2) {
      dp[v][1] = 0;
    } else if (f == 1) {
      long long x = b[poz];
      for (int i = 0; i < poz; ++i) {
        if (g[v][i] != p) x = (x * a[i]) % inf;
      }
      for (int i = poz + 1; i < a.size(); ++i) {
        if (g[v][i] != p) x = (x * a[i]) % inf;
      }
      dp[v][1] = x;
    } else {
      long long x = 1;
      for (int i = 0; i < a.size(); ++i) {
        x = (x * a[i]) % inf;
      }
      dp[v][1] = 0;
      for (int i = 0; i < a.size(); ++i) {
        dp[v][1] =
            (dp[v][1] + ((x * power(a[i], inf - 2)) % inf * b[i]) % inf) % inf;
      }
    }
  }
  dp[v][0] = dp[v][1];
  if (col[v] == 0) {
    long long x = 1;
    for (int i = 0; i < g[v].size(); ++i) {
      if (g[v][i] != p) {
        x = (x * dp[g[v][i]][0]) % inf;
      }
    }
    dp[v][0] = (dp[v][0] + x) % inf;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int p;
    scanf("%d", &p);
    g[i].push_back(p);
    g[p].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &col[i]);
  }
  dfs(0, -1);
  printf("%d\n", dp[0][1] % inf);
  return 0;
}
