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;
void exgcd(long long a, long long b, long long& d, long long& x, long long& y) {
  if (!b)
    d = a, x = 1LL, y = 0LL;
  else
    exgcd(b, a % b, d, y, x), y -= x * (a / b);
}
long long inv(long long a, long long m) {
  long long d, x, y;
  exgcd(a, m, d, x, y);
  return d == 1LL ? (x + m) % m : -1LL;
}
vector<int> vc[100010];
long long dp[100010][2], MOD = 1000000007;
int root, vis[100010], col[100010];
void dfs(int u) {
  int i, j, k, len = vc[u].size();
  if (col[u] == 1) {
    dp[u][1] = 1;
    for (i = 0; i < len; i++) {
      dfs(vc[u][i]);
      if (col[vc[u][i]] == 1)
        dp[u][1] *= dp[vc[u][i]][1];
      else
        dp[u][1] *= (dp[vc[u][i]][0] + dp[vc[u][i]][1]);
      dp[u][1] %= MOD;
    }
  } else if (col[u] == 0) {
    dp[u][0] = 1;
    for (i = 0; i < len; i++) {
      dfs(vc[u][i]);
      if (col[vc[u][i]] == 1)
        dp[u][0] *= dp[vc[u][i]][1];
      else
        dp[u][0] *= (dp[vc[u][i]][0] + dp[vc[u][i]][1]);
      dp[u][0] %= MOD;
    }
    for (i = 0; i < len; i++) {
      if (col[vc[u][i]] == 1)
        dp[u][1] +=
            dp[u][0] * inv(dp[vc[u][i]][1], MOD) % MOD * dp[vc[u][i]][1];
      else
        dp[u][1] += dp[u][0] * inv(dp[vc[u][i]][0] + dp[vc[u][i]][1], MOD) %
                    MOD * dp[vc[u][i]][1];
      dp[u][1] %= MOD;
    }
  }
  dp[u][1] %= MOD;
  dp[u][0] %= MOD;
}
void solve() {
  int T, t, n, m, i, j, k = 0, u;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d", &u);
    vc[u].push_back(i);
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &col[i]);
    k += col[i];
  }
  if (k == 0) {
    printf("0\n");
    return;
  }
  if (k == 1) {
    printf("1\n");
    return;
  }
  dfs(0);
  printf("%I64d\n", dp[0][1]);
}
int main() {
  solve();
  return 0;
}
