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;
int sgn(double x) { return (x > 1e-8) - (x < -1e-8); }
int count_bit(int x) { return x == 0 ? 0 : count_bit(x >> 1) + (x & 1); }
template <class T>
inline void ckmin(T &a, const T b) {
  if (b < a) a = b;
}
template <class T>
inline void ckmax(T &a, const T b) {
  if (b > a) a = b;
}
const int mod = 1000000007;
int mul(int x, int y) { return (long long)x * y % mod; }
int fmod(int a, int p) {
  int ret = 1;
  while (p) {
    if (p & 1) ret = mul(ret, a);
    a = mul(a, a);
    p >>= 1;
  }
  return ret;
}
int get_inv(int x) { return fmod(x, mod - 2); }
const int MAXN = 100000 + 10;
vector<int> g[MAXN];
int n, col[MAXN];
int dp[MAXN][2];
void dfs(int x, int f) {
  int allmul = 1;
  for (__typeof((g[x]).begin()) it = (g[x]).begin(); it != (g[x]).end(); ++it) {
    if (*it != f) {
      dfs(*it, x);
      allmul = mul(allmul, dp[*it][0] + dp[*it][1]);
    }
  }
  if (col[x]) {
    dp[x][1] = allmul;
  } else {
    dp[x][0] = allmul;
    for (__typeof((g[x]).begin()) it = (g[x]).begin(); it != (g[x]).end();
         ++it) {
      if (*it != f) {
        if (dp[*it][1]) {
          dp[x][1] +=
              mul(dp[*it][1], mul(allmul, get_inv(dp[*it][0] + dp[*it][1])));
          dp[x][1] %= mod;
        }
      }
    }
  }
}
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < (n); ++i) {
      g[i].clear();
    }
    for (int i = (1); i <= (n - 1); ++i) {
      int pa;
      scanf("%d", &pa);
      g[pa].push_back(i);
      g[i].push_back(pa);
    }
    for (int i = 0; i < (n); ++i) {
      scanf("%d", &col[i]);
    }
    memset(dp, 0, sizeof(dp));
    dfs(0, -1);
    printf("%d\n", dp[0][1]);
  }
  return 0;
}
