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;
long long const N = 2e5 + 20, MOD = 1e9 + 7;
long long n, x, dp[N][2];
bool vis[N], col[N];
vector<long long> child[N];
void dfs(int v = 0) {
  for (auto u : child[v]) dfs(u);
  vector<long long> preChild, sufChild;
  preChild.push_back(1), sufChild.push_back(1);
  for (auto u : child[v]) {
    preChild.push_back((preChild.back() * ((dp[u][0] + dp[u][1]) % MOD)) % MOD);
  }
  for (int i = child[v].size() - 1; i >= 0; i--) {
    int u = child[v][i];
    sufChild.push_back((sufChild.back() * ((dp[u][0] + dp[u][1]) % MOD)) % MOD);
  }
  if (col[v]) {
    dp[v][0] = 0;
    dp[v][1] = preChild.back();
    return;
  }
  dp[v][0] = preChild.back();
  dp[v][1] = 0;
  preChild.push_back(1), sufChild.push_back(1);
  for (int i = 0; i < child[v].size(); i++) {
    int u = child[v][i];
    dp[v][1] = (dp[v][1] + ((dp[u][1] * preChild[i] % MOD) *
                            (sufChild[sufChild.size() - 3 - i] % MOD) % MOD)) %
               MOD;
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> x;
    child[x].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) cin >> col[i];
  dfs();
  return cout << dp[0][1], 0;
}
