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 maxn = 111111;
const int MOD = 1e9 + 7;
int n;
int a[maxn];
struct Edge {
  int next, v;
} e[2 * maxn];
int en = 0;
int head[maxn];
void add(int x, int y) {
  e[en].v = y;
  e[en].next = head[x];
  head[x] = en++;
}
bool u[maxn];
long long f[maxn][2];
void dfs(int x) {
  int i;
  u[x] = 1;
  f[x][0] = 1;
  f[x][1] = 0;
  for (i = head[x]; i != -1; i = e[i].next) {
    if (!u[e[i].v]) {
      dfs(e[i].v);
      f[x][1] = (f[x][1] * f[e[i].v][0] + f[x][0] * f[e[i].v][1]) % MOD;
      f[x][0] = f[x][0] * f[e[i].v][0] % MOD;
    }
  }
  u[x] = 0;
  if (a[x] == 0)
    f[x][0] = (f[x][0] + f[x][1]) % MOD;
  else
    f[x][1] = f[x][0];
}
long long farm() {
  if (n == 1) return 1;
  memset(u, 0, sizeof(u));
  dfs(0);
  return f[0][1];
}
int main() {
  int i;
  int x;
  scanf("%d", &n);
  memset(head, -1, sizeof(head));
  en = 0;
  for (i = 0; i < (n - 1); i++) {
    scanf("%d", &x);
    add(i + 1, x);
    add(x, i + 1);
  }
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  printf("%I64d", farm());
  return 0;
}
