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;
bool checkbit(int x, int y) { return (x & (1 << y)); }
int setbit(int x, int y) { return (x ^ (1 << y)); }
const int dirs[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
const int mod = (1e9 + 7);
const int INF = (0x7fffffff);
const int maxn = 1e5 + 7;
vector<int> e[maxn];
bitset<maxn> c;
bitset<maxn> vis;
long long dp[maxn][2];
void dfs(int u) {
  vis[u] = 1;
  dp[u][1] = c[u];
  dp[u][0] = 1 - c[u];
  for (auto v : e[u]) {
    if (vis[v]) continue;
    dfs(v);
    long long old[2];
    old[0] = dp[u][0];
    old[1] = dp[u][1];
    dp[u][0] = (old[0] * dp[v][0]);
    dp[u][0] = (dp[u][0] + old[0] * dp[v][1]) % mod;
    dp[u][1] = (old[1] * dp[v][0]);
    dp[u][1] = (dp[u][1] + old[1] * dp[v][1]) % mod;
    dp[u][1] = (dp[u][1] + old[0] * dp[v][1]) % mod;
  }
}
int main() {
  vis.reset();
  int n;
  scanf("%d", &n);
  for (int i = (1); i < (n); i++) {
    int p;
    scanf("%d", &p);
    e[p].push_back(i);
  }
  for (int i = (0); i < (n); i++) {
    int clr;
    scanf("%d", &clr);
    c[i] = (clr == 1);
  }
  dfs(0);
  printf("%I64d\n", dp[0][1]);
  return 0;
}
