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 n, x, cnt;
bool col[200010];
long long dp[200010][2], p[200010], a[200010], b[200010];
vector<int> vs[200010];
void dfs(int u, int fa) {
  if (vs[u].size() == 1 && fa != 0) {
    if (col[u]) {
      dp[u][0] = 0;
      dp[u][1] = 1;
    } else {
      dp[u][0] = 1;
      dp[u][1] = 0;
    }
    return;
  }
  dp[u][0] = 1;
  dp[u][1] = 0;
  p[0] = 1;
  for (int i = 0; i < vs[u].size(); i++) {
    int v = vs[u][i];
    if (v == fa) continue;
    dfs(v, u);
    dp[u][0] *= dp[v][0] + dp[v][1];
    dp[u][0] %= 1000000007;
  }
  cnt = 0;
  for (int i = 0; i < vs[u].size(); i++) {
    int v = vs[u][i];
    if (v == fa) continue;
    b[++cnt] = v;
  }
  a[cnt + 1] = 1;
  if (col[u]) {
    dp[u][1] = dp[u][0];
    dp[u][0] = 0;
    return;
  }
  for (int i = 1; i <= cnt; i++) {
    p[i] = p[i - 1] * (dp[b[i]][0] + dp[b[i]][1]);
    p[i] %= 1000000007;
  }
  for (int i = cnt; i >= 1; i--) {
    a[i] = a[i + 1] * (dp[b[i]][0] + dp[b[i]][1]);
    a[i] %= 1000000007;
  }
  for (int i = 1; i <= cnt; i++) {
    dp[u][1] += dp[b[i]][1] * ((p[i - 1] * a[i + 1]) % 1000000007);
    dp[u][1] %= 1000000007;
  }
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &x);
    x++;
    vs[x].push_back(i);
    vs[i].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    cin >> col[i];
  }
  dfs(1, 0);
  cout << dp[1][1];
  return 0;
}
