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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int st[100100];
vector<int> adj[100100];
long long dp[100100][3];
void dfs(int u, int fa = 0) {
  if (st[u] == 1)
    dp[u][1] = 1;
  else
    dp[u][0] = 1;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v == fa) continue;
    dfs(v, u);
    dp[u][1] = dp[u][1] * (dp[v][0] + dp[v][1]) % 1000000007 +
               dp[u][0] * dp[v][1] % 1000000007;
    dp[u][1] %= 1000000007;
    dp[u][0] = (dp[u][0] * dp[v][0] + dp[u][0] * dp[v][1]) % 1000000007;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int p;
    scanf("%d", &p);
    adj[i].push_back(p + 1);
    adj[p + 1].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &st[i]);
  }
  dfs(1);
  cout << dp[1][1] << endl;
}
