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 LIM = 1e+5 + 10;
const int MOD = 1e+9 + 7;
int color[LIM];
long long cnt_1[LIM];
long long cnt_0[LIM];
int used[LIM];
vector<vector<int> > g(LIM, vector<int>());
void dfs(int v) {
  used[v] = true;
  if (color[v] == 1) {
    cnt_1[v] = 1;
    cnt_0[v] = 0;
  } else {
    cnt_1[v] = 0;
    cnt_0[v] = 1;
  }
  for (int i = 0; i < (int)g[v].size(); i++) {
    int to = g[v][i];
    if (used[to]) continue;
    dfs(to);
    long long t1 = 0;
    long long t0 = 0;
    t1 += cnt_1[v] * cnt_1[to];
    t0 += cnt_0[v] * cnt_1[to];
    t1 += cnt_1[v] * cnt_0[to];
    t1 += cnt_0[v] * cnt_1[to];
    t0 += cnt_0[v] * cnt_0[to];
    cnt_0[v] = t0 % MOD;
    cnt_1[v] = t1 % MOD;
  }
}
int main() {
  int n;
  cin >> n;
  int v = 0;
  for (int u = 1; u < n; u++) {
    cin >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < n; i++) cin >> color[i];
  for (int i = 0; i < n; i++)
    if (!used[i]) {
      dfs(i);
    }
  cout << cnt_1[0];
  return 0;
}
