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 maxn = 1e5 + 5;
const long long mod = 1e9 + 7;
vector<long long> AdjList[maxn];
long long dp[2][maxn];
long long col[maxn];
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long x = binpow(a, b / 2);
  x = (x * x) % mod;
  if (b % 2 == 0)
    return x;
  else
    return (x * a) % mod;
}
long long bindiv(long long a, long long b) {
  long long inv = binpow(b, mod - 2);
  return (a * inv) % mod;
}
void dfs(long long s, long long p) {
  for (long long u : AdjList[s]) {
    if (u != p) {
      dfs(u, s);
    }
  }
  if (col[s] == 1) {
    long long tmp1 = 1;
    long long tmp2 = 0;
    dp[0][s] = 0;
    for (long long u : AdjList[s]) {
      if (u != p) tmp1 = (tmp1 * dp[0][u] % mod + tmp1 * dp[1][u] % mod) % mod;
    }
    dp[1][s] = tmp1;
  } else {
    long long tmp1 = 1;
    long long tmp2 = 0;
    for (long long u : AdjList[s]) {
      if (u != p) {
        tmp2 = (tmp2 * dp[0][u] % mod + tmp2 * dp[1][u] % mod +
                tmp1 * dp[1][u] % mod) %
               mod;
        tmp1 = (tmp1 * dp[0][u] % mod + tmp1 * dp[1][u] % mod) % mod;
      }
    }
    dp[1][s] = tmp2;
    dp[0][s] = tmp1;
  }
}
signed main() {
  long long n;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long p;
    cin >> p;
    AdjList[p].push_back(i);
    AdjList[i].push_back(p);
  }
  for (long long i = 0; i < n; i++) {
    cin >> col[i];
    dp[0][i] = dp[1][i] = 0;
  }
  dfs(0, 0);
  cout << (dp[1][0]) % mod;
}
