<p>Given an integer array <code>arr</code> and an integer <code>k</code>, modify the array by repeating it <code>k</code> times.</p>

<p>For example, if <code>arr = [1, 2]</code> and <code>k = 3 </code>then the modified array will be <code>[1, 2, 1, 2, 1, 2]</code>.</p>

<p>Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be <code>0</code> and its sum in that case is <code>0</code>.</p>

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

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

<pre>
<strong>Input:</strong> arr = [1,2], k = 3
<strong>Output:</strong> 9
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,-2,1], k = 5
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> arr = [-1,-2], k = 7
<strong>Output:</strong> 0
</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;= k &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= arr[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def kConcatenationMaxSum(self, arr: List[int], k: int) -> int:
        #-100, 102, 99, -100, 102, 99
        #102, -220, 99, 102, -2200, 99, 102, -100, 99
        #1,-2,3
        #1,-1,2
        #3,1,2
        res = 0
        prefixForward = [arr[0]]
        prefixBackward = [arr[-1]]
        for i in range(1, len(arr)):
            prefixForward.append(prefixForward[i-1] + arr[i])
            prefixBackward.append(prefixBackward[i-1] + arr[len(arr)-1-i])
        x = max(prefixForward)
        y = max(prefixBackward)
        if sum(arr) <= 0 and k > 1:
            res = max(res, x + y)
        elif sum(arr) > 0 and k > 1:
            res = max(res, sum(arr) * (k - 2) + x + y)
        z = 0
        q = 0
        for i in range(len(arr)):
            if z + arr[i] < 0:
                z = 0
            else:
                z += arr[i]
            q = max(q, z)
        return max(res, x, y, q, max(arr)) % (10 ** 9 + 7)
30600