<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:
        # 分三种情况
        # k = 1: 同 53 题 注意 答案非负数
        # k = 2: 最大前缀和 + 最大后缀和， 其实两个合起来同(1)
        # k > 2: sum < 0, 同(2)；否则(2)答案 + sum * (k - 2)
        nums = arr * min(k, 2)
        ans = pre = 0 
        f = [0] * len(nums)
        f[0] = nums[0]
        for i in range(1, len(nums)):
            f[i] = max(f[i - 1], 0) + nums[i]
        ans = max(ans, max(f))
        s = sum(arr)
        if s > 0 and k > 2:
            ans = max(ans, s * (k - 2) + ans)
        return ans % 1000000007
        
        

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