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 nax = 100005, mod = 1e9 + 7;
vector<long long> g[nax];
long long c[nax], dp[nax][2] = {{0}};
long long mod_expo(long long a, long long b) {
  long long tans = 1;
  a = a % mod;
  while (b != 0) {
    if (b & 1) tans = (1LL * tans * a) % mod;
    a = (1LL * a * a) % mod;
    b /= 2;
  }
  return tans;
}
long long mod_inv(long long a) { return mod_expo(a, mod - 2); }
long long sum(long long a, long long b) { return (a + b) % mod; }
long long mul(long long a, long long b) { return (a * b) % mod; }
long long divm(long long a, long long b) { return (a * mod_inv(b)) % mod; }
void dfs(long long u, long long p) {
  for (auto &k : g[u])
    if (k != p) dfs(k, u);
  if (c[u] == 0) {
    dp[u][0] = 1;
    for (auto &k : g[u])
      if (k != p) dp[u][0] = (dp[u][0] * (dp[k][0] + dp[k][1])) % mod;
    dp[u][1] = 0;
    long long zpi = 1;
    for (auto &k : g[u])
      if (k != p) zpi = (zpi * (dp[k][0] + dp[k][1])) % mod;
    for (auto &k : g[u])
      if (k != p)
        dp[u][1] =
            sum(dp[u][1], mul(divm(zpi, (dp[k][0] + dp[k][1])), dp[k][1]));
  } else {
    dp[u][0] = 0;
    dp[u][1] = 1;
    for (auto &k : g[u])
      if (k != p) dp[u][1] = (dp[u][1] * (dp[k][0] + dp[k][1])) % mod;
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long x;
    cin >> x;
    g[i + 1].push_back(x);
    g[x].push_back(i + 1);
  }
  for (long long i = 0; i < n; i++) cin >> c[i];
  dfs(0, 0);
  cout << dp[0][1] << endl;
}
