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;
struct edge {
  int v, nxt;
};
int n;
const int MOD = 1000000007;
edge e[200020];
int g[100010];
int nume;
int color[100010];
long long f[100010][2];
void addedge(int u, int v) {
  e[++nume].v = v;
  e[nume].nxt = g[u];
  g[u] = nume;
  e[++nume].v = u;
  e[nume].nxt = g[v];
  g[v] = nume;
}
void init() {
  int x, i;
  nume = 0;
  memset(g, 0, sizeof(g));
  for (i = 1; i < n; i++) {
    scanf("%d", &x);
    addedge(x, i);
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &color[i]);
  }
}
void dfs(int x, int fa) {
  int i;
  int y;
  if (color[x] == 1) {
    f[x][1] = 1;
    f[x][0] = 0;
  } else {
    f[x][1] = 0;
    f[x][0] = 1;
  }
  for (i = g[x]; i; i = e[i].nxt)
    if (e[i].v != fa) {
      y = e[i].v;
      dfs(y, x);
      f[x][1] = (f[x][1] * (f[y][0] + f[y][1]) + f[x][0] * f[y][1]) % MOD;
      f[x][0] = f[x][0] * (f[y][0] + f[y][1]) % MOD;
    }
}
int main() {
  scanf("%d", &n);
  init();
  dfs(0, -1);
  printf("%I64d\n", f[0][1]);
  return 0;
}
