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;
vector<int> g[100010];
bitset<100010> coler;
long long ans;
const int mod = 1e9 + 7;
long long d[100010][2];
void dfs(int x, int p) {
  for (int i = 0; i < g[x].size(); i++) {
    if (g[x][i] != p) dfs(g[x][i], x);
  }
  d[x][coler.test(x)] = 1;
  for (int i = 0; i < g[x].size(); i++) {
    if (g[x][i] == p) continue;
    d[x][1] = d[g[x][i]][1] * d[x][0] + d[g[x][i]][0] * d[x][1];
    d[x][0] = d[g[x][i]][0] * d[x][0];
    d[x][0] %= mod;
    d[x][1] %= mod;
  }
  d[x][0] += d[x][1];
  d[x][0] %= mod;
  d[x][1] %= mod;
  return;
}
int main() {
  int n;
  while (cin >> n) {
    ans = 1;
    coler.reset();
    memset(d, 0, sizeof(d));
    for (int i = 0; i < n; i++) g[i].clear();
    int v;
    for (int i = 1; i < n; i++) {
      scanf("%d", &v);
      g[i].push_back(v);
      g[v].push_back(i);
    }
    for (int i = 0; i < n; i++) {
      scanf("%d", &v);
      if (v == 1) {
        coler.set(i);
      }
    }
    dfs(0, 0);
    cout << d[0][1] << endl;
  }
  return 0;
}
