<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) {
        int n = arr.size();
        long long sum =0, maxSum =0, mod = 1e9+7;
        for(int i= 0; i<n;i++){
            sum += arr[i];
            maxSum = max(maxSum,sum);
            if(sum <0){
                sum = 0;
            }
        }
        if(k ==1){
            return maxSum % mod;
        }
        else{
            vector<int> doubleArr = arr;
            doubleArr.insert(doubleArr.end(),arr.begin(),arr.end());
            sum =0; maxSum =0;
            for(int i=0; i<doubleArr.size(); i++){
                sum += doubleArr[i];
                maxSum = max(maxSum,sum);
                if(sum <0)
                    sum =0;
            }

            long long total =0;
            for(int i=0; i<n; i++){
                total += arr[i];
            }
            long long ans = maxSum + ((k-2)*total);
            if(total >0){
                return ans%mod;
            }
            return maxSum%mod;
        }
    }
};
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