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
n = int(input())
edges = [int(x) for x in input().split()]
color = [int(x) for x in input().split()]
graph = [[] for _ in range(n)]

for a,b in enumerate(edges):
    graph[a+1].append(b)
    graph[b].append(a+1)
dp = [[0]*2 for _ in range(n)]
visited = [0]*n

stack = [0]
while stack:
    v = stack[-1]
    visited[v] = -1
    cn = 0
    for u in graph[v]:
        if visited[u] is not 0:
            continue
        else:
            cn += 1
            stack.append(u)
    if not cn:
        dp[v][0] = 1
        dp[v][1] = 0
        for u in graph[v]:
            if visited[u] is -1:
                continue
            dp[v][1] *= dp[u][0]
            dp[v][1] += dp[v][0] * dp[u][1]
            dp[v][0] *= dp[u][0]
            dp[v][1] %= 1000000007
            dp[v][0] %= 1000000007
            
        if color[v] is 1:
            dp[v][1] = dp[v][0]
        else:
            dp[v][0] += dp[v][1]
            dp[v][0] %= 1000000007
        visited[v] = 1
        stack.pop()
        

# def dfs(v):
    # dp[v][0] = 1
    # dp[v][1] = 0
#     visited[v] = True
#     for u in graph[v]:
        # if visited[u] is True:
        #     continue
#         dfs(u)
#         dp[v][1] *= dp[u][0]
#         dp[v][1] += dp[v][0] * dp[u][1]
#         dp[v][0] *= dp[u][0]
#         dp[v][1] %= 1000000007
#         dp[v][0] %= 1000000007
        
#     if color[v] is 1:
#         dp[v][1] = dp[v][0]
#     else:
#         dp[v][0] += dp[v][1]
#         dp[v][0] %= 1000000007

# dfs(0)
ans = dp[0][1]
print(ans)
