<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 {
public:
    int kConcatenationMaxSum(vector<int>& arr, int k) {
        const int MOD = 1e9 + 7;
        int n = arr.size();
        
        int onesum = 0, currentSum = 0;
        for (int i = 0; i < n; ++i) {
            currentSum = max(arr[i], currentSum + arr[i]);
            onesum = max(onesum, currentSum);
        }
        
        if (k == 1) return onesum % MOD;
        
        long long totalSum = accumulate(arr.begin(), arr.end(), 0LL);
        
        int twosum = 0;
        currentSum = 0;
        for (int i = 0; i < 2 * n; ++i) {
            currentSum = max(arr[i % n], currentSum + arr[i % n]);
            twosum = max(twosum, currentSum);
        }
        
        if(k>1){
            if (totalSum > 0) {
                return max((long long)twosum, onesum + (k - 1) * totalSum) % MOD;
            } 
            else {
                return max((long long)twosum, (long long)onesum) % MOD;
            }
        }
        return onesum;
    }
};

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