<p>Given an integer array&nbsp;<code>nums</code>, return <em>the number of longest increasing subsequences.</em></p>

<p><strong>Notice</strong> that the sequence has to be <strong>strictly</strong> increasing.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,5,4,7]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,2,2,2]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The length of the longest increasing subsequence is 1, and there are 5 increasing subsequences of length 1, so output 5.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2000</code></li>
	<li><code>-10<sup>6</sup> &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li>The answer is guaranteed to fit inside a 32-bit integer.</li>
</ul>

class Solution:
    def findNumberOfLIS(self, nums: List[int]) -> int:

        
        maxl_dp= [1]*len(nums)
        count_dp= [1]*len(nums)


        for i in range(len(nums)):
            for j in range(i):
                if nums[i]>nums[j]:
                    if maxl_dp[i]<1+maxl_dp[j]:
                        maxl_dp[i]=1+maxl_dp[j]
                        count_dp[i]=count_dp[j]
                    elif maxl_dp[i]==1+maxl_dp[j]:
                        count_dp[i]+=count_dp[j]
                    
        maxl= max(maxl_dp)

        res=0
        for i in range(len(nums)):
            if maxl_dp[i]==maxl:
                res+=count_dp[i]
        return res


16600