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 int maxn = 100010;
const int mod = 1000000007;
long long dp[maxn][2];
vector<int> G[maxn];
int col[maxn];
void gcd(long long a, long long b, long long &d, long long &x, long long &y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    gcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long niyuan(long long a) {
  long long x, y, d;
  gcd(a, mod, d, x, y);
  return (x + mod) % mod;
}
void dfs(int u, int fa) {
  if (G[u].size() == 0) {
    dp[u][col[u]] = 1;
    return;
  }
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    dfs(v, u);
  }
  if (col[u]) {
    dp[u][1] = 1;
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      dp[u][1] *= dp[v][0] + dp[v][1];
      dp[u][1] %= mod;
    }
    if (dp[u][1] == 0) dp[u][1] = 1;
  } else {
    dp[u][0] = 1;
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      dp[u][0] *= dp[v][0] + dp[v][1];
      dp[u][0] %= mod;
    }
    long long tmp = dp[u][0];
    dp[u][1] = 0;
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (dp[v][1]) {
        long long t = (tmp * niyuan(dp[v][0] + dp[v][1])) % mod;
        dp[u][1] += dp[v][1] * t;
        dp[u][1] %= mod;
      }
    }
  }
}
int main() {
  int n, pi;
  while (cin >> n) {
    for (int i = 0; i < n; i++) G[i].clear();
    for (int i = 1; i < n; i++) {
      scanf("%d", &pi);
      G[pi].push_back(i);
    }
    for (int i = 0; i < n; i++) scanf("%d", col + i);
    memset(dp, 0, sizeof dp);
    dfs(0, -1);
    cout << dp[0][1] << endl;
  }
  return 0;
}
