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 long long MOD = 1000000007;
void dfs(int akt, int oj, vector<vector<int> >& T, vector<int>& col,
         vector<vector<long long> >& dp) {
  for (int i = 0; i < T[akt].size(); ++i)
    if (T[akt][i] != oj) dfs(T[akt][i], akt, T, col, dp);
  if (T[akt].size() == 1 && akt != 0) {
    if (col[akt] == 0) {
      dp[akt][0] = 1;
      dp[akt][1] = 0;
    } else {
      dp[akt][0] = 0;
      dp[akt][1] = 1;
    }
  } else {
    int sa = T[akt].size();
    vector<long long> pref(sa), suf(sa);
    if (T[akt][0] == oj)
      pref[0] = 1;
    else
      pref[0] = dp[T[akt][0]][0] + dp[T[akt][0]][1];
    for (int i = 1; i < sa; ++i) {
      pref[i] = pref[i - 1];
      if (T[akt][i] != oj)
        pref[i] = (pref[i] * (dp[T[akt][i]][0] + dp[T[akt][i]][1])) % MOD;
    }
    if (T[akt][sa - 1] == oj)
      suf[sa - 1] = 1;
    else
      suf[sa - 1] = dp[T[akt][sa - 1]][0] + dp[T[akt][sa - 1]][1];
    for (int i = sa - 2; i >= 0; --i) {
      suf[i] = suf[i + 1];
      if (T[akt][i] != oj)
        suf[i] = (suf[i] * (dp[T[akt][i]][0] + dp[T[akt][i]][1])) % MOD;
    }
    if (col[akt] == 0) {
      dp[akt][1] = 0;
      for (int i = 0; i < sa; ++i) {
        if (T[akt][i] != oj) {
          long long resz = 1;
          if (i != 0) resz *= pref[i - 1];
          if (i != sa - 1) resz *= suf[i + 1];
          resz %= MOD;
          resz *= dp[T[akt][i]][1];
          resz %= MOD;
          dp[akt][1] = (dp[akt][1] + resz) % MOD;
        }
      }
      dp[akt][0] = pref[sa - 1];
    } else {
      dp[akt][1] = pref[sa - 1];
      dp[akt][0] = 0;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<vector<int> > T(n);
  for (int i = 0; i < n - 1; ++i) {
    int p;
    cin >> p;
    T[i + 1].push_back(p);
    T[p].push_back(i + 1);
  }
  vector<int> col(n);
  for (int i = 0; i < n; ++i) cin >> col[i];
  vector<vector<long long> > dp(n, vector<long long>(2));
  dfs(0, -1, T, col, dp);
  cout << dp[0][1];
  return 0;
}
