<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 {
    long M = 1e9+7;
public:
    int kConcatenationMaxSum(vector<int>& arr, int k) 
    {   
        if (k==1)
            return maxSubArrSum(arr)%M; 
        
        long arrSum = 0; 
        for (int i = 0; i < arr.size(); i++) 
        {
            arrSum += arr[i]; 
            arrSum %= M;
        }            
         
        int len = arr.size();
        for (int i=0; i<len; i++)
            arr.push_back(arr[i]);
        
        if (arrSum < 0)                     
            return maxSubArrSum(arr)%M;         
        else        
            return (maxSubArrSum(arr) + ((long)k - 2) * arrSum)%M;               
    }
    
    long maxSubArrSum(vector<int>arr) 
    { 
        long max_so_far = 0; 
        long max_ending_here = 0; 
  
        for (int i = 0; i < arr.size(); i++) 
        { 
            max_ending_here = max_ending_here + arr[i]; 
            if (max_ending_here < 0) 
                max_ending_here = 0; 
            if (max_so_far < max_ending_here) 
                max_so_far = max_ending_here;
        } 
        return max_so_far; 
    } 
};

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