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 = 100005;
const long long mod = 1e9 + 7;
vector<int> edge[Max];
int is[Max];
long long dp[Max][2];
long long inv(long long x) {
  long long res = 1;
  for (int k = mod - 2; k; k >>= 1) {
    if (k & 1) res = res * x % mod;
    x = x * x % mod;
  }
  return res;
}
void dfs(int u, int f) {
  int v, m = edge[u].size();
  long long tot = 1;
  long long tmp;
  for (int i = 0; i < m; i++) {
    v = edge[u][i];
    if (v == f) continue;
    dfs(v, u);
    tot = tot * (dp[v][0] + dp[v][1]) % mod;
  }
  if (is[u])
    dp[u][0] = 0;
  else
    dp[u][0] = tot;
  if (is[u])
    dp[u][1] = tot;
  else {
    dp[u][1] = 0;
    for (int i = 0; i < m; i++) {
      v = edge[u][i];
      if (v == f) continue;
      tmp = (tot * dp[v][1] % mod) * inv(dp[v][0] + dp[v][1]) % mod;
      dp[u][1] = (dp[u][1] + tmp) % mod;
    }
  }
}
int main() {
  int n, p;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d", &p);
    edge[i].push_back(p);
    edge[p].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &is[i]);
  }
  dfs(0, -1);
  cout << dp[0][1] << endl;
  return 0;
}
