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 N = 100005;
const long long MOD = 1000000007;
int n, node[N];
vector<int> g[N];
long long dp[N][2];
long long pow_mod(long long x, long long k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * x % MOD;
    x = x * x % MOD;
    k >>= 1;
  }
  return ans;
}
long long inv(long long x) { return pow_mod(x, MOD - 2); }
void init() {
  scanf("%d", &n);
  int u;
  for (int i = 1; i < n; i++) {
    scanf("%d", &u);
    g[u].push_back(i);
  }
  for (int i = 0; i < n; i++) scanf("%d", &node[i]);
}
void dfs(int u) {
  if (g[u].size() == 0) {
    dp[u][node[u]] = 1;
    return;
  }
  for (int i = 0; i < g[u].size(); i++) dfs(g[u][i]);
  dp[u][0] = dp[u][1] = 1;
  if (node[u]) {
    dp[u][0] = 0;
    for (int i = 0; i < g[u].size(); i++) {
      int v = g[u][i];
      dp[u][1] = dp[u][1] * (dp[v][0] + dp[v][1]) % MOD;
    }
  } else {
    long long cnt = 0;
    long long mul = 1;
    for (int i = 0; i < g[u].size(); i++) {
      int v = g[u][i];
      dp[u][0] = dp[u][0] * (dp[v][0] + dp[v][1]) % MOD;
      mul = mul * (dp[v][0] + dp[v][1]) % MOD;
    }
    dp[u][1] = 0;
    for (int i = 0; i < g[u].size(); i++) {
      int v = g[u][i];
      dp[u][1] =
          (dp[u][1] + mul * inv((dp[v][0] + dp[v][1]) % MOD) % MOD * dp[v][1]) %
          MOD;
    }
  }
}
int main() {
  init();
  dfs(0);
  printf("%I64d\n", dp[0][1] % MOD);
  return 0;
}
