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;
long long power(long long x, long long y) {
  long long res = 1LL;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
long long inv(long long n) { return power(n, 1000000007 - 2) % 1000000007; }
long long isprime(long long n) {
  if (n < 2) return 0;
  long long i;
  for (i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
long long n, x, col[200005], dp[2][200005];
vector<long long> a[200005];
void dfs(long long v, long long p) {
  long long leaf = 1;
  long long pro = 1;
  for (auto x : a[v]) {
    if (x == p) continue;
    dfs(x, v);
    pro *= (dp[0][x] + dp[1][x]) % 1000000007;
    pro %= 1000000007;
  }
  if (leaf) {
    if (col[v]) {
      dp[1][v] = 1;
      dp[0][v] = 0;
    } else {
      dp[1][v] = 0;
      dp[0][v] = 1;
    }
  }
  if (col[v]) {
    dp[1][v] = pro;
    dp[0][v] = 0;
  } else {
    dp[0][v] = pro;
    for (auto x : a[v]) {
      if (x == p) continue;
      dp[1][v] +=
          ((pro * inv((dp[0][x] + dp[1][x]) % 1000000007) % 1000000007) *
           dp[1][x]) %
          1000000007;
      dp[1][v] %= 1000000007;
    }
  }
}
void solve() {
  cin >> n;
  long long i;
  for (i = 1; i < n; i++) {
    cin >> x;
    a[i].push_back(x);
    a[x].push_back(i);
  }
  for (i = 0; i < n; i++) cin >> col[i];
  dfs(0, -1);
  cout << dp[1][0] << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  long long tc = 0;
  while (t--) {
    tc++;
    solve();
  }
}
