<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>

const long long MOD = 1e9 + 7;

class Solution {
public:
    int kConcatenationMaxSum(vector<int>& arr, int k) {
        
        int n = arr.size();
        long long sum = 0;


        long long curr = 0;
        long long maxSub = 0;

        for(int i = 0; i < n; i++) {
            curr += arr[i];
            maxSub = max(maxSub, curr);
            if(curr < 0) {
                curr = 0;
            }
            sum += arr[i];
        }

        curr = 0;

        long long maxSubTwo = 0;
        for(int i = 0; i < 2*n; i++) {
            curr += arr[i % n];
            maxSubTwo = max(maxSubTwo, curr);
            if(curr < 0) {
                curr = 0;
            }
        }

        if(k == 1) {
            return maxSub;
        } else if(k == 2) {
            return maxSubTwo % MOD;
        } else {
            if(sum >= 0) {
                long long val1 = ( (sum % MOD ) * ((k - 2) * 1LL % MOD) ) % MOD;
                long long val2 = maxSubTwo % MOD;
                long long finalSum = (val1 + val2) % MOD;
                return finalSum;
            } else {
                return maxSubTwo;
            }
        }

        
    }
};
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