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 maxn = 2 * 100 * 1000 + 100;
const long long delta = 1000 * 1000 * 1000 + 7;
int n, col[maxn];
vector<int> adj[maxn];
long long dp[2][maxn];
bool dfsmark[maxn];
void inp();
void dfs(int v);
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  inp();
  dfs(0);
  cout << dp[1][0] << '\n';
  return 0;
}
void inp() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u = i + 1, v;
    cin >> v;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  for (int i = 0; i < n; i++) cin >> col[i];
}
void dfs(int v) {
  dfsmark[v] = true;
  vector<int> cnt;
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (!dfsmark[u]) {
      dfs(u);
      cnt.push_back(u);
    }
  }
  if (!cnt.size()) {
    dp[0][v] = !col[v];
    dp[1][v] = col[v];
    return;
  }
  if (col[v]) {
    dp[0][v] = 0;
    long long res1 = 1;
    for (int i = 0; i < cnt.size(); i++) {
      int u = cnt[i];
      res1 = (res1 * (dp[0][u] + dp[1][u])) % delta;
    }
    dp[1][v] = res1;
    return;
  }
  long long res0 = 1;
  for (int i = 0; i < cnt.size(); i++) {
    int u = cnt[i];
    res0 = (res0 * (dp[0][u] + dp[1][u])) % delta;
  }
  dp[0][v] = res0;
  long long p = 1, res1 = 0;
  for (int i = 0; i < cnt.size(); i++) {
    int u = cnt[i];
    long long tmp = 0;
    tmp = (res1 * (dp[0][u] + dp[1][u])) % delta;
    tmp = (tmp + p * dp[1][u]) % delta;
    res1 = tmp;
    p = (p * (dp[0][u] + dp[1][u])) % delta;
  }
  dp[1][v] = res1;
}
