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 MOD = 1000000007;
const int N = 100000;
int n, root;
vector<vector<int> > a;
bool is_dist[N];
long long dp[N][2];
void read_input() {
  scanf("%d ", &n);
  a.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int p;
    scanf("%d ", &p);
    a[i + 1].push_back(p);
    a[p].push_back(i + 1);
  }
  root = -1;
  for (int i = 0; i < n; i++) {
    int is_dist_int;
    scanf("%d ", &is_dist_int);
    is_dist[i] = is_dist_int == 1;
    if (root == -1 && is_dist[i]) root = i;
  }
}
long long binary_exp(long long x, long long m) {
  if (m == 0)
    return 1;
  else if (m % 2 == 1)
    return (binary_exp(x, m - 1) * x) % MOD;
  else {
    long long val = binary_exp(x, m / 2);
    return (val * val) % MOD;
  }
}
inline long long mod_inv(long long x) { return binary_exp(x, MOD - 2); }
void dfs(int u, int parent) {
  if (is_dist[u]) {
    dp[u][0] = 0;
    dp[u][1] = 1;
    for (int v : a[u]) {
      if (v == parent) continue;
      dfs(v, u);
      dp[u][1] = (dp[u][1] * ((dp[v][0] + dp[v][1]) % MOD)) % MOD;
    }
  } else {
    dp[u][0] = 1;
    dp[u][1] = 0;
    long long zero_mul = 1;
    for (int v : a[u]) {
      if (v == parent) continue;
      dfs(v, u);
      dp[u][0] = (dp[u][0] * (dp[v][0] + dp[v][1])) % MOD;
      zero_mul = (zero_mul * (dp[v][0] + dp[v][1])) % MOD;
    }
    for (int v : a[u]) {
      if (v == parent) continue;
      dp[u][1] =
          (dp[u][1] +
           (zero_mul * ((dp[v][1] * mod_inv(dp[v][0] + dp[v][1])) % MOD)) %
               MOD) %
          MOD;
    }
  }
}
void solve() {
  if (root == -1) {
    printf("0\n");
    return;
  }
  dfs(root, -1);
  printf("%lld\n", dp[root][1]);
}
int main() {
  read_input();
  solve();
  return 0;
}
