<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) {
        long long MOD=1e9+7;
        long long sum=accumulate(arr.begin(), arr.end(), 0);
        int n =arr.size();
        vector<long long> prefix(n+1, 0), suffix(n+1, 0);
        for(int i=0;i<n;i++) prefix[i+1]+=arr[i]+prefix[i];
        for(int i=n-1;i>=0;i--) suffix[i]+=arr[i]+suffix[i+1];
        long long maxSuffix = *max_element(suffix.begin(), suffix.end());
        long long maxPrefix = *max_element(prefix.begin(), prefix.end());
        if(k==1) return maxSuffix>maxPrefix?maxSuffix:maxPrefix;
        long long maxSub=0, sumt=0;
        int start=0;
        for(int i=0;i<n;i++){
            sumt=(sumt+arr[i])%MOD;
            while(sumt<0 && start<=i){
              sumt-=arr[start];
              start++;
            }
            if(sumt> maxSub){
                maxSub=sumt;
            }
        }
        if(sum>0){
            long maxisum=(maxPrefix+maxSuffix+sum*(k-2));
            return maxisum>maxSub?maxisum%MOD:maxSub%MOD;
        }
        return maxSuffix+maxPrefix>maxSub?maxSuffix+maxPrefix:maxSub;
        // return -1;
    }
};
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