# Time:  O(n^2)
# Space: O(n)

# Given an array of unique integers, each integer is strictly greater than 1.
# We make a binary tree using these integers and each number may be used for
# any number of times.
# Each non-leaf node's value should be equal to the product of the values of
# it's children.
# How many binary trees can we make?  Return the answer modulo 10 ** 9 + 7.
#
# Example 1:
#
# Input: A = [2, 4]
# Output: 3
# Explanation: We can make these trees: [2], [4], [4, 2, 2]
# Example 2:
#
# Input: A = [2, 4, 5, 10]
# Output: 7
# Explanation: We can make these trees:
#              [2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2].
#
# Note:
# - 1 <= A.length <= 1000.
# - 2 <= A[i] <= 10 ^ 9.

try:
    xrange          # Python 2
except NameError:
    xrange = range  # Python 3


class Solution(object):
    def numFactoredBinaryTrees(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        M = 10**9 + 7
        A.sort()
        dp = {}
        for i in xrange(len(A)):
            dp[A[i]] = 1
            for j in xrange(i):
                if A[i] % A[j] == 0 and A[i] // A[j] in dp:
                    dp[A[i]] += dp[A[j]] * dp[A[i] // A[j]]
                    dp[A[i]] %= M
        return sum(dp.values()) % M

# Time:  O(n^2)
# Space: O(n)

# Given an array of unique integers, each integer is strictly greater than 1.
# We make a binary tree using these integers and each number may be used for
# any number of times.
# Each non-leaf node's value should be equal to the product of the values of
# it's children.
# How many binary trees can we make?  Return the answer modulo 10 ** 9 + 7.
#
# Example 1:
#
# Input: A = [2, 4]
# Output: 3
# Explanation: We can make these trees: [2], [4], [4, 2, 2]
# Example 2:
#
# Input: A = [2, 4, 5, 10]
# Output: 7
# Explanation: We can make these trees:
#              [2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2].
#
# Note:
# - 1 <= A.length <= 1000.
# - 2 <= A[i] <= 10 ^ 9.

try:
    xrange          # Python 2
except NameError:
    xrange = range  # Python 3


class Solution(object):
    def numFactoredBinaryTrees(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        M = 10**9 + 7
        A.sort()
        dp = {}
        for i in xrange(len(A)):
            dp[A[i]] = 1
            for j in xrange(i):
                if A[i] % A[j] == 0 and A[i] // A[j] in dp:
                    dp[A[i]] += dp[A[j]] * dp[A[i] // A[j]]
                    dp[A[i]] %= M
        return sum(dp.values()) % M

# Time:  O(n^2) # Space: O(n) # Given an array of unique integers, each integer is strictly greater than 1. # We make a binary tree using these integers and each number may be used for # any number of times. # Each non-leaf node's value should be equal to the product of the values of # it's children. # How many binary trees can we make?  Return the answer modulo 10 ** 9 + 7. # # Example 1: # # Input: A = [2, 4] # Output: 3 # Explanation: We can make these trees: [2], [4], [4, 2, 2] # Example 2: # # Input: A = [2, 4, 5, 10] # Output: 7 # Explanation: We can make these trees: #              [2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2]. # # Note: # - 1 <= A.length <= 1000. # - 2 <= A[i] <= 10 ^ 9. # Python 2 # Python 3 :type A: List[int]         :rtype: int