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>
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
using namespace std;
const long long INF = 1e10;
struct Edge {
  long long v = 0, u = 0, q = 0;
  Edge() {}
  Edge(int v1, int u1) {
    v = v1;
    u = u1;
  }
  bool operator<(const Edge &p) const { return tie(v, u) < tie(p.v, p.u); }
};
vector<vector<int>> g;
vector<int> a;
vector<int> used;
vector<Edge> dp;
const long long MOD = 1e9 + 7;
void dfs(int v) {
  used[v] = 1;
  dp[v].v = 1;
  for (auto it : g[v]) {
    if (!used[it]) {
      dfs(it);
      dp[v].u = ((dp[v].u * dp[it].v) % MOD + (dp[v].v * dp[it].u) % MOD) % MOD;
      dp[v].v = (dp[v].v * dp[it].v) % MOD;
    }
  }
  if (a[v] == 1)
    dp[v].u = dp[v].v;
  else
    dp[v].v = (dp[v].v + dp[v].u) % MOD;
}
int main() {
  int n;
  cin >> n;
  g.resize(n);
  a.resize(n);
  dp.assign(n, Edge(0, 0));
  used.assign(n, 0);
  for (int i = 0; i < n - 1; ++i) {
    int t;
    cin >> t;
    g[t].push_back(i + 1);
  }
  for (int i = 0; i < n; ++i) cin >> a[i];
  dfs(0);
  cout << dp[0].u;
}
