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;
int n;
vector<vector<int> > sou;
vector<int> bl;
vector<long long> c, b;
void dfs(int v, int f = -1);
int main() {
  scanf("%d", &n);
  sou.resize(n);
  bl.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int p;
    scanf("%d", &p);
    sou[i + 1].push_back(p);
    sou[p].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) scanf("%d", &bl[i]);
  c.resize(n);
  b.resize(n);
  dfs(0);
  printf("%d\n", (int)(c[0]));
  return 0;
}
void euc(long long a, long long b, long long &x, long long &y) {
  if (a < b) {
    euc(b, a, y, x);
    return;
  }
  if (b == 0) {
    x = 1;
    y = 0;
    return;
  }
  long long xx, yy;
  euc(b, a % b, xx, yy);
  x = yy;
  y = xx - (a / b) * yy;
  while (y < 0) {
    x -= b;
    y += a;
  }
}
void dfs(int v, int f) {
  if (sou[v].size() == 1 && sou[v][0] == f) {
    if (bl[v]) {
      c[v] = 1;
      b[v] = 0;
    } else {
      c[v] = 0;
      b[v] = 1;
    }
  } else {
    if (bl[v]) {
      b[v] = 0;
      c[v] = 1;
      for (int i = 0; i < sou[v].size(); i++) {
        int s = sou[v][i];
        if (s == f) continue;
        dfs(s, v);
        c[v] = (c[v] * (b[s] + c[s])) % mod;
      }
    } else {
      b[v] = 1;
      c[v] = 0;
      long long prod = 1;
      for (int i = 0; i < sou[v].size(); i++) {
        int s = sou[v][i];
        if (s == f) continue;
        dfs(s, v);
        prod = (prod * (b[s] + c[s])) % mod;
      }
      for (int i = 0; i < sou[v].size(); i++) {
        int s = sou[v][i];
        if (s == f) continue;
        b[v] = (b[v] * (b[s] + c[s])) % mod;
        long long t, trash;
        euc((b[s] + c[s]) % mod, mod, t, trash);
        t = (t * prod) % mod;
        t = (t * c[s]) % mod;
        c[v] = (c[v] + t) % mod;
      }
    }
  }
}
