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>
const double EPS = 1e-24;
const long long int MOD = 1000000007ll;
const double PI = 3.14159265359;
int INF = 2147483645;
long long int INFINF = 9223372036854775807;
template <class T>
T Max2(T a, T b) {
  return a < b ? b : a;
}
template <class T>
T Min2(T a, T b) {
  return a < b ? a : b;
}
template <class T>
T Max3(T a, T b, T c) {
  return Max2(Max2(a, b), c);
}
template <class T>
T Min3(T a, T b, T c) {
  return Min2(Min2(a, b), c);
}
template <class T>
T Max4(T a, T b, T c, T d) {
  return Max2(Max2(a, b), Max2(c, d));
}
template <class T>
T Min4(T a, T b, T c, T d) {
  return Min2(Min2(a, b), Max2(c, d));
}
using namespace std;
long long int N;
vector<long long int> G[100010];
long long int dp[100010][2];
bool vis[100010];
long long int color[100010];
void dfs(long long int u) {
  vis[u] = true;
  if (color[u]) {
    dp[u][0] = 0;
    dp[u][1] = 1;
  } else {
    dp[u][0] = 1;
    dp[u][1] = 0;
  }
  for (long long int v : G[u]) {
    if (!vis[v]) {
      dfs(v);
      long long int temp[2];
      temp[0] = dp[u][0];
      temp[1] = dp[u][0];
      temp[1] = (dp[u][1] * dp[v][1]) % MOD;
      temp[1] = (temp[1] + dp[u][0] * dp[v][1]) % MOD;
      temp[1] = (temp[1] + dp[u][1] * dp[v][0]) % MOD;
      temp[0] = (dp[u][0] * dp[v][1]) % MOD;
      temp[0] = (temp[0] + dp[u][0] * dp[v][0]) % MOD;
      dp[u][0] = temp[0];
      dp[u][1] = temp[1];
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N;
  for (int i = 1; i <= N - 1; i++) {
    long long int x;
    cin >> x;
    G[x].push_back(i);
    G[i].push_back(x);
  }
  for (int i = 0; i < N; i++) cin >> color[i];
  dfs(0);
  cout << dp[0][1] << "\n";
  return 0;
}
