<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 kadanesalgo(vector<int> arr){
        int sum=0;
        int maxsum = INT_MIN;
        int n=arr.size();
        for(int i=0;i<n;i++){
            sum = sum + arr[i];
            if(sum<0){
                sum=0;
                maxsum = max(maxsum,arr[i]);
            }
            else{
                maxsum = max(maxsum,sum);
            }
        }
        return maxsum;
    }
    int kadanesalgoOf2array(vector<int> arr){
        int sum=0;
        int maxsum = INT_MIN;
        int n=arr.size();
        for(int i=0;i<n*2;i++){
            int idx=i%n;
            sum = sum + arr[idx];
            if(sum<0){
                sum=0;
                maxsum = max(maxsum,arr[idx]);
            }
            else{
                maxsum = max(maxsum,sum);
            }
        }
        return maxsum;
    }
    int kConcatenationMaxSum(vector<int>& arr, int k) {
        long long arrsum=0;int MOD=1e9+7;
        for(auto num:arr)arrsum+=num;
        long long ans=0;
        if(k==1)ans=kadanesalgo(arr)%MOD;
        else if(arrsum<0)ans=kadanesalgoOf2array(arr)%MOD;
        else
            ans=(kadanesalgoOf2array(arr)+(k-2)*arrsum)%MOD;

        return ans>0?ans:0;    
    }
};
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