<p>Given an array of integers <code>arr</code>, return <em>the number of subarrays with an <strong>odd</strong> sum</em>.</p>

<p>Since the answer can be very large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [1,3,5]
<strong>Output:</strong> 4
<strong>Explanation:</strong> All subarrays are [[1],[1,3],[1,3,5],[3],[3,5],[5]]
All sub-arrays sum are [1,4,9,3,8,5].
Odd sums are [1,9,3,5] so the answer is 4.
</pre>

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

<pre>
<strong>Input:</strong> arr = [2,4,6]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All subarrays are [[2],[2,4],[2,4,6],[4],[4,6],[6]]
All sub-arrays sum are [2,6,12,4,10,6].
All sub-arrays have even sum and the answer is 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,2,3,4,5,6,7]
<strong>Output:</strong> 16
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 100</code></li>
</ul>

class Solution:
    def numOfSubarrays(self, arr: List[int]) -> int:
        # wow
        even = odd = ans = 0

        for i in range(1, len(arr)):
            arr[i] += arr[i-1]
                
        for n in arr:
            if n % 2:
                ans = (ans + 1 + even) % (1e9+7)
                odd += 1
            else:
                even += 1
                ans = (ans + odd) % (1e9+7)
                
        return int(ans)

            
#         dp = [i % 2 == 1 for i in arr]
        
#         count = sum(dp)
        
#         for i in range(1, len(arr)):
#             temp = list(dp)
#             for j in range(i, len(arr)):
#                 new = (arr[j] % 2 == 1 + dp[j-1]) == 1
#                 count = (count + int(new)) % (1e9+7)
#                 temp[j] = new
#             dp = temp
            
#         return int(count)
24000